Functions of Several Variables and Partial Derivatives ¶ While there are some subtleties involved in moving from univariate to multivariate calculus, much of the intuition carries over.
Consider a function
y = f ( x 1 , x 2 , … , x n ) . y = f(x_1, x_2, \ldots, x_n). y = f ( x 1 , x 2 , … , x n ) . The partial derivative of y y y i i i x i x_i x i
∂ y ∂ x i = lim Δ x i → 0 f ( x 1 , … , x i + Δ x i , … , x n ) − f ( x 1 , … , x i , … , x n ) Δ x i \frac{\partial y}{\partial x_i}
= \lim_{\Delta x_i \to 0}
\frac{f(x_1, \ldots, x_i + \Delta x_i, \ldots, x_n)
- f(x_1, \ldots, x_i, \ldots, x_n)}{\Delta x_i} ∂ x i ∂ y = Δ x i → 0 lim Δ x i f ( x 1 , … , x i + Δ x i , … , x n ) − f ( x 1 , … , x i , … , x n ) provided the limit exists.
Common notations include
∂ y ∂ x i , ∂ ∂ x i f ( x 1 , … , x n ) , f i . \frac{\partial y}{\partial x_i},
\quad
\frac{\partial}{\partial x_i} f(x_1,\ldots,x_n),
\quad
f_i. ∂ x i ∂ y , ∂ x i ∂ f ( x 1 , … , x n ) , f i . For a function f ( x , y ) f(x,y) f ( x , y ) x x x f x ( x , y ) f_x(x,y) f x ( x , y ) f x f_x f x f ( x 1 , x 2 ) f(x_1,x_2) f ( x 1 , x 2 ) x 1 x_1 x 1 f 1 ( x , y ) f_1(x,y) f 1 ( x , y ) f 1 f_1 f 1
Rules of Partial Differentiation ¶ The rules of partial differentiation are the same as those for univariate calculus, except that all other variables are treated as constants .
Examples
a) Let
f ( x 1 , x 2 ) = x 1 + 3 x 2 2 . f(x_1,x_2) = \sqrt{x_1 + 3x_2^2}. f ( x 1 , x 2 ) = x 1 + 3 x 2 2 . has two partial derivatives:
f 1 = 1 2 x 1 + 3 x 2 2 , f 2 = 3 x 2 x 1 + 3 x 2 2 . f_1 = \frac{1}{2\sqrt{x_1 + 3x_2^2}},
\qquad
f_2 = \frac{3x_2}{\sqrt{x_1 + 3x_2^2}}. f 1 = 2 x 1 + 3 x 2 2 1 , f 2 = x 1 + 3 x 2 2 3 x 2 . b) Similarly, let
h ( x , y , z ) = ln ( 5 x + 2 y − 3 z ) . h(x,y,z) = \ln(5x + 2y - 3z). h ( x , y , z ) = ln ( 5 x + 2 y − 3 z ) . then
h x = 5 5 x + 2 y − 3 z , h y = 2 5 x + 2 y − 3 z , h z = − 3 5 x + 2 y − 3 z . h_x = \frac{5}{5x+2y-3z}, \quad
h_y = \frac{2}{5x+2y-3z}, \quad
h_z = \frac{-3}{5x+2y-3z}. h x = 5 x + 2 y − 3 z 5 , h y = 5 x + 2 y − 3 z 2 , h z = 5 x + 2 y − 3 z − 3 . Example - Cobb-Douglass production function ¶ Consider the Cobb–Douglas production function
Q = A K 1 − α L α . Q = AK^{1-\alpha}L^\alpha. Q = A K 1 − α L α . The marginal product of labor (MPL) can be found by taking the partial derivative of output with respect to labor:
∂ Q ∂ L = α A K 1 − α L α − 1 = α A ( K L ) 1 − α \frac{\partial Q}{\partial L}
= \alpha A K^{1 - \alpha} L^{\alpha - 1}
= \alpha A \left( \frac{K}{L} \right)^{1 - \alpha} ∂ L ∂ Q = α A K 1 − α L α − 1 = α A ( L K ) 1 − α In neo-classical economics theory, workers are paid a real wage equal to their marginal productivity of labor. So in the equation above we see that the real wage and MPL increase with larger capital stock relative to labor.
Similarly,
∂ Q ∂ K = ( 1 − α ) A K − α L α = ( 1 − α ) A ( L K ) α \frac{\partial Q}{\partial K}
= (1 - \alpha) A K^{-\alpha} L^{\alpha}
= (1 - \alpha) A \left( \frac{L}{K} \right)^{\alpha} ∂ K ∂ Q = ( 1 − α ) A K − α L α = ( 1 − α ) A ( K L ) α Which suggests that the rate of return to capital is relatively high in countries that have a relatively large labor-capital ratios. These high rates of returns should cause capital inflow from rich to poor countries. But why do we not see this in reality? See Lucas (1990)
Multivariate Derivative Rules ¶ (1) Product Rule
∂ ∂ x [ f ( x , y ) g ( x , y ) ] = g ( x , y ) ∂ f ( x , y ) ∂ x + f ( x , y ) ∂ g ( x , y ) ∂ x , ∂ ∂ y [ f ( x , y ) g ( x , y ) ] = g ( x , y ) ∂ f ( x , y ) ∂ y + f ( x , y ) ∂ g ( x , y ) ∂ y . \begin{aligned}
\frac{\partial}{\partial x}[f(x,y)g(x,y)]
&=
g(x,y)\frac{\partial f(x,y)}{\partial x}
+
f(x,y)\frac{\partial g(x,y)}{\partial x}, \\[2em]
\frac{\partial}{\partial y}[f(x,y)g(x,y)]
&=
g(x,y)\frac{\partial f(x,y)}{\partial y}
+
f(x,y)\frac{\partial g(x,y)}{\partial y}.
\end{aligned} ∂ x ∂ [ f ( x , y ) g ( x , y )] ∂ y ∂ [ f ( x , y ) g ( x , y )] = g ( x , y ) ∂ x ∂ f ( x , y ) + f ( x , y ) ∂ x ∂ g ( x , y ) , = g ( x , y ) ∂ y ∂ f ( x , y ) + f ( x , y ) ∂ y ∂ g ( x , y ) . (2) Quotient Rule
The derivative of the quotient of two (differentiable) functions, f ( x , y ) / g ( x , y ) f(x,y)/g(x,y) f ( x , y ) / g ( x , y )
∂ ∂ x [ f ( x , y ) g ( x , y ) ] = g ( x , y ) ∂ f ( x , y ) ∂ x − f ( x , y ) ∂ g ( x , y ) ∂ x [ g ( x , y ) ] 2 , ∂ ∂ y [ f ( x , y ) g ( x , y ) ] = g ( x , y ) ∂ f ( x , y ) ∂ y − f ( x , y ) ∂ g ( x , y ) ∂ y [ g ( x , y ) ] 2 . \begin{aligned}
\frac{\partial}{\partial x}\left[\frac{f(x,y)}{g(x,y)}\right]
&=
\frac{
g(x,y)\frac{\partial f(x,y)}{\partial x}
-
f(x,y)\frac{\partial g(x,y)}{\partial x}
}{
[g(x,y)]^2
}, \\[2em]
\frac{\partial}{\partial y}\left[\frac{f(x,y)}{g(x,y)}\right]
&=
\frac{
g(x,y)\frac{\partial f(x,y)}{\partial y}
-
f(x,y)\frac{\partial g(x,y)}{\partial y}
}{
[g(x,y)]^2
}.
\end{aligned} ∂ x ∂ [ g ( x , y ) f ( x , y ) ] ∂ y ∂ [ g ( x , y ) f ( x , y ) ] = [ g ( x , y ) ] 2 g ( x , y ) ∂ x ∂ f ( x , y ) − f ( x , y ) ∂ x ∂ g ( x , y ) , = [ g ( x , y ) ] 2 g ( x , y ) ∂ y ∂ f ( x , y ) − f ( x , y ) ∂ y ∂ g ( x , y ) . where f ( x , y ) f(x,y) f ( x , y ) g ( x , y ) g(x,y) g ( x , y ) g ( x , y ) ≠ 0 g(x,y)\neq 0 g ( x , y ) = 0
(3) Generalized Power Function
∂ [ g ( x , y ) ] n ∂ x = n [ g ( x , y ) ] n − 1 ∂ g ( x , y ) ∂ x , ∂ [ g ( x , y ) ] n ∂ y = n [ g ( x , y ) ] n − 1 ∂ g ( x , y ) ∂ y . \begin{aligned}
\frac{\partial [g(x,y)]^n}{\partial x}
&=
n[g(x,y)]^{\,n-1}\frac{\partial g(x,y)}{\partial x}, \\[1em]
\frac{\partial [g(x,y)]^n}{\partial y}
&=
n[g(x,y)]^{\,n-1}\frac{\partial g(x,y)}{\partial y}.
\end{aligned} ∂ x ∂ [ g ( x , y ) ] n ∂ y ∂ [ g ( x , y ) ] n = n [ g ( x , y ) ] n − 1 ∂ x ∂ g ( x , y ) , = n [ g ( x , y ) ] n − 1 ∂ y ∂ g ( x , y ) . where n n n g ( x , y ) g(x,y) g ( x , y )
Second-Order and Cross-Partial Derivatives ¶ For
y = f ( x 1 , x 2 ) , y = f(x_1,x_2), y = f ( x 1 , x 2 ) , the second partial derivatives are
f 11 = ∂ 2 y ∂ x 1 2 , f 22 = ∂ 2 y ∂ x 2 2 . f_{11} = \frac{\partial^2 y}{\partial x_1^2},
\qquad
f_{22} = \frac{\partial^2 y}{\partial x_2^2}. f 11 = ∂ x 1 2 ∂ 2 y , f 22 = ∂ x 2 2 ∂ 2 y . The cross partial derivatives are
f 12 = ∂ 2 y ∂ x 1 ∂ x 2 , f 21 = ∂ 2 y ∂ x 2 ∂ x 1 . f_{12} = \frac{\partial^2 y}{\partial x_1 \partial x_2},
\qquad
f_{21} = \frac{\partial^2 y}{\partial x_2 \partial x_1}. f 12 = ∂ x 1 ∂ x 2 ∂ 2 y , f 21 = ∂ x 2 ∂ x 1 ∂ 2 y . Continuing with the examples above, the function
f ( x 1 , x 2 ) = x 1 + 3 x 2 2 f(x_1, x_2) = \sqrt{x_1 + 3x_2^2} f ( x 1 , x 2 ) = x 1 + 3 x 2 2 has two second partial derivatives
f 11 ( x 1 , x 2 ) = − 1 4 ( x 1 + 3 x 2 2 ) 3 / 2 , f 22 ( x 1 , x 2 ) = − 9 x 2 2 ( x 1 + 3 x 2 2 ) 3 / 2 + 3 ( x 1 + 3 x 2 2 ) 1 / 2 \begin{aligned}
f_{11}(x_1,x_2)
&=
-\frac{1}{4\left(x_1 + 3x_2^2\right)^{3/2}}, \\[2em]
f_{22}(x_1,x_2)
&=
-\frac{9x_2^2}{\left(x_1 + 3x_2^2\right)^{3/2}}
+
\frac{3}{\left(x_1 + 3x_2^2\right)^{1/2}}
\end{aligned} f 11 ( x 1 , x 2 ) f 22 ( x 1 , x 2 ) = − 4 ( x 1 + 3 x 2 2 ) 3/2 1 , = − ( x 1 + 3 x 2 2 ) 3/2 9 x 2 2 + ( x 1 + 3 x 2 2 ) 1/2 3 and the cross partial derivative
f 12 ( x 1 , x 2 ) = f 21 ( x 1 , x 2 ) = − 3 x 2 2 ( x 1 + 3 x 2 2 ) 3 / 2 . f_{12}(x_1,x_2)
=
f_{21}(x_1,x_2)
=
-\frac{3x_2}{2\left(x_1 + 3x_2^2\right)^{3/2}}. f 12 ( x 1 , x 2 ) = f 21 ( x 1 , x 2 ) = − 2 ( x 1 + 3 x 2 2 ) 3/2 3 x 2 . For the function
h ( x , y , z ) = ln ( 5 x + 2 y − 3 z ) h(x,y,z) = \ln(5x + 2y - 3z) h ( x , y , z ) = ln ( 5 x + 2 y − 3 z ) There are three second partial derivatives:
h x x ( x , y , z ) = − 25 ( 5 x + 2 y − 3 z ) 2 , h_{xx}(x,y,z)
=
-\frac{25}{(5x + 2y - 3z)^2}, h xx ( x , y , z ) = − ( 5 x + 2 y − 3 z ) 2 25 , h y y ( x , y , z ) = − 4 ( 5 x + 2 y − 3 z ) 2 , and h z z ( x , y , z ) = − 9 ( 5 x + 2 y − 3 z ) 2 . h_{yy}(x,y,z)
=
-\frac{4}{(5x + 2y - 3z)^2},
\quad \text{and} \quad
h_{zz}(x,y,z)
=
-\frac{9}{(5x + 2y - 3z)^2}. h yy ( x , y , z ) = − ( 5 x + 2 y − 3 z ) 2 4 , and h zz ( x , y , z ) = − ( 5 x + 2 y − 3 z ) 2 9 . The function also has three cross partial derivatives:
h x y ( x , y , z ) = h y x ( x , y , z ) = − 10 ( 5 x + 2 y − 3 z ) 2 , h_{xy}(x,y,z)
=
h_{yx}(x,y,z)
=
-\frac{10}{(5x + 2y - 3z)^2}, h x y ( x , y , z ) = h y x ( x , y , z ) = − ( 5 x + 2 y − 3 z ) 2 10 , h x z ( x , y , z ) = h z x ( x , y , z ) = − 15 ( 5 x + 2 y − 3 z ) 2 , and h_{xz}(x,y,z)
=
h_{zx}(x,y,z)
=
-\frac{15}{(5x + 2y - 3z)^2},
\quad \text{and} \quad h x z ( x , y , z ) = h z x ( x , y , z ) = − ( 5 x + 2 y − 3 z ) 2 15 , and h y z ( x , y , z ) = h z y ( x , y , z ) = − 6 ( 5 x + 2 y − 3 z ) 2 h_{yz}(x,y,z)
=
h_{zy}(x,y,z)
=
-\frac{6}{(5x + 2y - 3z)^2} h yz ( x , y , z ) = h zy ( x , y , z ) = − ( 5 x + 2 y − 3 z ) 2 6 Young’s Theorem ¶ If all second-order partial derivatives are continuous, then
f i j = f j i . f_{ij} = f_{ji}. f ij = f ji . More generally, if all the partial derivatives of the function
f ( x 1 , x 2 , … , x n ) f(x_1, x_2, \ldots, x_n) f ( x 1 , x 2 , … , x n ) exist and are themselves differentiable with continuous derivatives, then
∂ ∂ x i ( ∂ f ( x 1 , x 2 , … , x n ) ∂ x j ) = ∂ ∂ x j ( ∂ f ( x 1 , x 2 , … , x n ) ∂ x i ) . \frac{\partial}{\partial x_i}
\left(
\frac{\partial f(x_1, x_2, \ldots, x_n)}{\partial x_j}
\right)
=
\frac{\partial}{\partial x_j}
\left(
\frac{\partial f(x_1, x_2, \ldots, x_n)}{\partial x_i}
\right). ∂ x i ∂ ( ∂ x j ∂ f ( x 1 , x 2 , … , x n ) ) = ∂ x j ∂ ( ∂ x i ∂ f ( x 1 , x 2 , … , x n ) ) . Or, written differently,
f j i ( x 1 , x 2 , … , x n ) = f i j ( x 1 , x 2 , … , x n ) , f_{ji}(x_1, x_2, \ldots, x_n)
=
f_{ij}(x_1, x_2, \ldots, x_n), f ji ( x 1 , x 2 , … , x n ) = f ij ( x 1 , x 2 , … , x n ) , for any i i i j j j n n n
Young’s Theorem shows that a multivariate function that is fully differentiable with respect to all its n n n n n n ( n 2 − n ) / 2 (n^2 - n)/2 ( n 2 − n ) /2
For our Cobb–Douglas production function,
Q = A K 1 − α L α , Q = AK^{1-\alpha}L^{\alpha}, Q = A K 1 − α L α , the second partial derivative with respect to L L L K K K
∂ 2 Q ∂ L 2 = − ( 1 − α ) α A K 1 − α L α − 1 , \frac{\partial^2 Q}{\partial L^2}
=
-(1-\alpha)\alpha A K^{1-\alpha}L^{\alpha-1}, ∂ L 2 ∂ 2 Q = − ( 1 − α ) α A K 1 − α L α − 1 , ∂ 2 Q ∂ K 2 = − ( 1 − α ) α A K − α − 1 L α . \frac{\partial^2 Q}{\partial K^2}
=
-(1-\alpha)\alpha A K^{-\alpha-1}L^{\alpha}. ∂ K 2 ∂ 2 Q = − ( 1 − α ) α A K − α − 1 L α . Each of these second partial derivatives is negative, reflecting diminishing
marginal productivity.
The single cross partial derivative is
∂ 2 Q ∂ K ∂ L = ∂ 2 Q ∂ L ∂ K = ( 1 − α ) α A K − α L α − 1 , \frac{\partial^2 Q}{\partial K \partial L}
=
\frac{\partial^2 Q}{\partial L \partial K}
=
(1-\alpha)\alpha A K^{-\alpha}L^{\alpha-1}, ∂ K ∂ L ∂ 2 Q = ∂ L ∂ K ∂ 2 Q = ( 1 − α ) α A K − α L α − 1 , which is positive since K K K L L L
The second-order derivatives of the Cobb–Douglas production function can be
summarized by the Hessian matrix
H ( Q ) = [ ∂ 2 Q ∂ K 2 ∂ 2 Q ∂ K ∂ L ∂ 2 Q ∂ L ∂ K ∂ 2 Q ∂ L 2 ] H(Q)
=
\begin{bmatrix}
\displaystyle \frac{\partial^2 Q}{\partial K^2}
&
\displaystyle \frac{\partial^2 Q}{\partial K \partial L}
\\[1em]
\displaystyle \frac{\partial^2 Q}{\partial L \partial K}
&
\displaystyle \frac{\partial^2 Q}{\partial L^2}
\end{bmatrix} H ( Q ) = ⎣ ⎡ ∂ K 2 ∂ 2 Q ∂ L ∂ K ∂ 2 Q ∂ K ∂ L ∂ 2 Q ∂ L 2 ∂ 2 Q ⎦ ⎤ The signs of the Hessian entries reflect diminishing marginal productivity
(negative diagonal terms) and complementarity between inputs (positive
off-diagonal terms).
For each of the following functions:
Compute the first-order partial derivatives
∂ y ∂ x 1 , ∂ y ∂ x 2 . \frac{\partial y}{\partial x_1}, \qquad \frac{\partial y}{\partial x_2}. ∂ x 1 ∂ y , ∂ x 2 ∂ y . For each functions, evaluate the first-order partial derivatives at the point
x 1 = 1 , x 2 = 4. x_1 = 1, \quad x_2 = 4. x 1 = 1 , x 2 = 4. For each functions, compute the second-order partial derivatives
∂ 2 y ∂ x 1 2 , ∂ 2 y ∂ x 2 2 , \frac{\partial^2 y}{\partial x_1^2}, \qquad
\frac{\partial^2 y}{\partial x_2^2}, ∂ x 1 2 ∂ 2 y , ∂ x 2 2 ∂ 2 y , and the cross partial derivative
∂ 2 y ∂ x 1 ∂ x 2 . \frac{\partial^2 y}{\partial x_1 \partial x_2}. ∂ x 1 ∂ x 2 ∂ 2 y . Assume all functions are defined on domains where the expressions are well-defined.
(i) y = f ( x 1 , x 2 ) = 12 x 1 4 − 6 x 1 2 x 2 + 4 x 2 3 y = f(x_1,x_2) = 12x_1^4 - 6x_1^2x_2 + 4x_2^3 y = f ( x 1 , x 2 ) = 12 x 1 4 − 6 x 1 2 x 2 + 4 x 2 3
(ii) y = f ( x 1 , x 2 ) = ( 3 x 1 2 + 5 x 2 + 1 ) ( x 2 + 4 ) y = f(x_1,x_2) = (3x_1^2 + 5x_2 + 1)(x_2 + 4) y = f ( x 1 , x 2 ) = ( 3 x 1 2 + 5 x 2 + 1 ) ( x 2 + 4 )
(iii) y = f ( x 1 , x 2 ) = 7 x 1 − x 1 x 2 2 x 1 − 2 y = f(x_1,x_2) = \frac{7x_1 - x_1x_2^2}{x_1 - 2} y = f ( x 1 , x 2 ) = x 1 − 2 7 x 1 − x 1 x 2 2
(iv) y = f ( x 1 , x 2 ) = ( 2 e x 1 ) ( e 2 x 1 x 2 2 ) y = f(x_1,x_2) = (2e^{x_1})(e^{2x_1}x_2^2) y = f ( x 1 , x 2 ) = ( 2 e x 1 ) ( e 2 x 1 x 2 2 )
(v) y = f ( x 1 , x 2 ) = 2 ln ( 3 x 1 ) − 4 ln ( 2 x 1 x 2 ) y = f(x_1,x_2) = 2\ln(3x_1) - 4\ln(2x_1x_2) y = f ( x 1 , x 2 ) = 2 ln ( 3 x 1 ) − 4 ln ( 2 x 1 x 2 )
(vi) y = f ( x 1 , x 2 ) = x 1 2 + 2 x 1 x 2 1 / 2 − 4 x 2 y = f(x_1,x_2) = x_1^2 + 2x_1x_2^{1/2} - 4x_2 y = f ( x 1 , x 2 ) = x 1 2 + 2 x 1 x 2 1/2 − 4 x 2
Answers: First-Order Partial Derivatives
(i) ∂ y ∂ x 1 = 48 x 1 3 − 12 x 1 x 2 , ∂ y ∂ x 2 = − 6 x 1 2 + 12 x 2 2 \frac{\partial y}{\partial x_1} = 48x_1^3 - 12x_1x_2,
\qquad
\frac{\partial y}{\partial x_2} = -6x_1^2 + 12x_2^2 ∂ x 1 ∂ y = 48 x 1 3 − 12 x 1 x 2 , ∂ x 2 ∂ y = − 6 x 1 2 + 12 x 2 2
(ii) ∂ y ∂ x 1 = 6 x 1 x 2 + 24 x 1 , ∂ y ∂ x 2 = 3 x 1 2 + 5 x 1 + 1 \frac{\partial y}{\partial x_1} = 6x_1x_2 + 24x_1,
\qquad
\frac{\partial y}{\partial x_2} = 3x_1^2 + 5x_1 + 1 ∂ x 1 ∂ y = 6 x 1 x 2 + 24 x 1 , ∂ x 2 ∂ y = 3 x 1 2 + 5 x 1 + 1
(iii) ∂ y ∂ x 1 = 2 x 2 2 − 14 ( x 1 − 2 ) 2 , ∂ y ∂ x 2 = − 2 x 1 x 2 x 1 − 2 \frac{\partial y}{\partial x_1} = \frac{2x_2^2 - 14}{(x_1 - 2)^2},
\qquad
\frac{\partial y}{\partial x_2} =
-\frac{2x_1x_2}{x_1 - 2} ∂ x 1 ∂ y = ( x 1 − 2 ) 2 2 x 2 2 − 14 , ∂ x 2 ∂ y = − x 1 − 2 2 x 1 x 2
(iv) ∂ y ∂ x 1 = 6 x 2 2 e 3 x 1 , ∂ y ∂ x 2 = 4 x 2 e 3 x 1 \frac{\partial y}{\partial x_1} = 6x_2^2 e^{3x_1},
\qquad
\frac{\partial y}{\partial x_2} = 4x_2 e^{3x_1} ∂ x 1 ∂ y = 6 x 2 2 e 3 x 1 , ∂ x 2 ∂ y = 4 x 2 e 3 x 1
(v) ∂ y ∂ x 1 = − 2 x 1 , ∂ y ∂ x 2 = − 4 x 2
\frac{\partial y}{\partial x_1} = -\frac{2}{x_1},
\qquad
\frac{\partial y}{\partial x_2} = -\frac{4}{x_2} ∂ x 1 ∂ y = − x 1 2 , ∂ x 2 ∂ y = − x 2 4
(vi) ∂ y ∂ x 1 = 2 x 1 + 2 x 2 , ∂ y ∂ x 2 = x 1 x 2 − 4 \frac{\partial y}{\partial x_1} = 2x_1 + 2\sqrt{x_2},
\qquad
\frac{\partial y}{\partial x_2} = \frac{x_1}{\sqrt{x_2}} - 4 ∂ x 1 ∂ y = 2 x 1 + 2 x 2 , ∂ x 2 ∂ y = x 2 x 1 − 4
Answers 2
(i) ∂ y ∂ x 1 ∣ ( 1 , 4 ) = 0 , ∂ y ∂ x 2 ∣ ( 1 , 4 ) = 186 \left.\frac{\partial y}{\partial x_1}\right|_{(1,4)} = 0,
\qquad
\left.\frac{\partial y}{\partial x_2}\right|_{(1,4)} = 186 ∂ x 1 ∂ y ∣ ∣ ( 1 , 4 ) = 0 , ∂ x 2 ∂ y ∣ ∣ ( 1 , 4 ) = 186
(ii) ∂ y ∂ x 1 ∣ ( 1 , 4 ) = 88 , ∂ y ∂ x 2 ∣ ( 1 , 4 ) = 9 \left.\frac{\partial y}{\partial x_1}\right|_{(1,4)} = 88,
\qquad
\left.\frac{\partial y}{\partial x_2}\right|_{(1,4)} = 9 ∂ x 1 ∂ y ∣ ∣ ( 1 , 4 ) = 88 , ∂ x 2 ∂ y ∣ ∣ ( 1 , 4 ) = 9
(iii) ∂ y ∂ x 1 ∣ ( 1 , 4 ) = 18 , ∂ y ∂ x 2 ∣ ( 1 , 4 ) = 8 \left.\frac{\partial y}{\partial x_1}\right|_{(1,4)} = 18,
\qquad
\left.\frac{\partial y}{\partial x_2}\right|_{(1,4)} = 8 ∂ x 1 ∂ y ∣ ∣ ( 1 , 4 ) = 18 , ∂ x 2 ∂ y ∣ ∣ ( 1 , 4 ) = 8
(iv) ∂ y ∂ x 1 ∣ ( 1 , 4 ) = 1928 , ∂ y ∂ x 2 ∣ ( 1 , 4 ) = 321 \left.\frac{\partial y}{\partial x_1}\right|_{(1,4)} = 1928,
\qquad
\left.\frac{\partial y}{\partial x_2}\right|_{(1,4)} = 321 ∂ x 1 ∂ y ∣ ∣ ( 1 , 4 ) = 1928 , ∂ x 2 ∂ y ∣ ∣ ( 1 , 4 ) = 321
(v) ∂ y ∂ x 1 ∣ ( 1 , 4 ) = − 2 , ∂ y ∂ x 2 ∣ ( 1 , 4 ) = − 1 \left.\frac{\partial y}{\partial x_1}\right|_{(1,4)} = -2,
\qquad
\left.\frac{\partial y}{\partial x_2}\right|_{(1,4)} = -1 ∂ x 1 ∂ y ∣ ∣ ( 1 , 4 ) = − 2 , ∂ x 2 ∂ y ∣ ∣ ( 1 , 4 ) = − 1
(vi) ∂ y ∂ x 1 ∣ ( 1 , 4 ) = 6 , ∂ y ∂ x 2 ∣ ( 1 , 4 ) = − 7 2 \left.\frac{\partial y}{\partial x_1}\right|_{(1,4)} = 6,
\qquad
\left.\frac{\partial y}{\partial x_2}\right|_{(1,4)} = -\frac{7}{2} ∂ x 1 ∂ y ∣ ∣ ( 1 , 4 ) = 6 , ∂ x 2 ∂ y ∣ ∣ ( 1 , 4 ) = − 2 7
3. Second-Order and Cross Partial Derivatives
For each function, compute: ∂ 2 y ∂ x 1 2 \dfrac{\partial^2 y}{\partial x_1^2} ∂ x 1 2 ∂ 2 y ∂ 2 y ∂ x 2 2 \dfrac{\partial^2 y}{\partial x_2^2} ∂ x 2 2 ∂ 2 y ∂ 2 y ∂ x 1 ∂ x 2 \dfrac{\partial^2 y}{\partial x_1\partial x_2} ∂ x 1 ∂ x 2 ∂ 2 y
Answers: Second-Order and Cross Partial Derivatives
(i) ∂ 2 y ∂ x 1 2 = 144 x 1 2 − 12 x 2 , ∂ 2 y ∂ x 2 2 = 24 x 2 , ∂ 2 y ∂ x 1 ∂ x 2 = − 12 x 1
\frac{\partial^2 y}{\partial x_1^2} = 144x_1^2 - 12x_2,
\qquad
\frac{\partial^2 y}{\partial x_2^2} = 24x_2,
\qquad
\frac{\partial^2 y}{\partial x_1\partial x_2} = -12x_1 ∂ x 1 2 ∂ 2 y = 144 x 1 2 − 12 x 2 , ∂ x 2 2 ∂ 2 y = 24 x 2 , ∂ x 1 ∂ x 2 ∂ 2 y = − 12 x 1
(ii) ∂ 2 y ∂ x 1 2 = 6 x 2 + 24 , ∂ 2 y ∂ x 2 2 = 6 x 1 + 5 , ∂ 2 y ∂ x 1 ∂ x 2 = 6 x 1
\frac{\partial^2 y}{\partial x_1^2} = 6x_2 + 24,
\qquad
\frac{\partial^2 y}{\partial x_2^2} = 6x_1 + 5,
\qquad
\frac{\partial^2 y}{\partial x_1\partial x_2} = 6x_1 ∂ x 1 2 ∂ 2 y = 6 x 2 + 24 , ∂ x 2 2 ∂ 2 y = 6 x 1 + 5 , ∂ x 1 ∂ x 2 ∂ 2 y = 6 x 1
(iii) ∂ 2 y ∂ x 1 2 = − 4 ( x 1 x 2 2 − 2 x 2 2 − 7 x 1 + 14 ) ( x 1 − 2 ) 3 , ∂ 2 y ∂ x 2 2 = 4 x 1 ( x 1 − 2 ) 2 , ∂ 2 y ∂ x 1 ∂ x 2 = − 4 x 2 ( x 1 − 2 ) 2
\frac{\partial^2 y}{\partial x_1^2} =
-\frac{4(x_1x_2^2 - 2x_2^2 - 7x_1 + 14)}{(x_1 - 2)^3},
\qquad
\frac{\partial^2 y}{\partial x_2^2} =
\frac{4x_1}{(x_1 - 2)^2},
\qquad
\frac{\partial^2 y}{\partial x_1\partial x_2} =
-\frac{4x_2}{(x_1 - 2)^2} ∂ x 1 2 ∂ 2 y = − ( x 1 − 2 ) 3 4 ( x 1 x 2 2 − 2 x 2 2 − 7 x 1 + 14 ) , ∂ x 2 2 ∂ 2 y = ( x 1 − 2 ) 2 4 x 1 , ∂ x 1 ∂ x 2 ∂ 2 y = − ( x 1 − 2 ) 2 4 x 2
(iv) ∂ 2 y ∂ x 1 2 = 18 x 2 2 e 3 x 1 , ∂ 2 y ∂ x 2 2 = 4 e 3 x 1 , ∂ 2 y ∂ x 1 ∂ x 2 = 12 e 3 x 1 x 2
\frac{\partial^2 y}{\partial x_1^2} = 18x_2^2 e^{3x_1},
\qquad
\frac{\partial^2 y}{\partial x_2^2} = 4e^{3x_1},
\qquad
\frac{\partial^2 y}{\partial x_1\partial x_2} = 12e^{3x_1}x_2 ∂ x 1 2 ∂ 2 y = 18 x 2 2 e 3 x 1 , ∂ x 2 2 ∂ 2 y = 4 e 3 x 1 , ∂ x 1 ∂ x 2 ∂ 2 y = 12 e 3 x 1 x 2
(v) ∂ 2 y ∂ x 1 2 = 2 x 1 2 , ∂ 2 y ∂ x 2 2 = 4 x 2 2 , ∂ 2 y ∂ x 1 ∂ x 2 = 0
\frac{\partial^2 y}{\partial x_1^2} = \frac{2}{x_1^2},
\qquad
\frac{\partial^2 y}{\partial x_2^2} = \frac{4}{x_2^2},
\qquad
\frac{\partial^2 y}{\partial x_1\partial x_2} = 0 ∂ x 1 2 ∂ 2 y = x 1 2 2 , ∂ x 2 2 ∂ 2 y = x 2 2 4 , ∂ x 1 ∂ x 2 ∂ 2 y = 0
(vi) ∂ 2 y ∂ x 1 2 = 2 , ∂ 2 y ∂ x 2 2 = − 1 2 x 1 x 2 − 3 / 2 , ∂ 2 y ∂ x 1 ∂ x 2 = x 2 − 1 / 2
\frac{\partial^2 y}{\partial x_1^2} = 2,
\qquad
\frac{\partial^2 y}{\partial x_2^2} = -\frac{1}{2}x_1 x_2^{-3/2},
\qquad
\frac{\partial^2 y}{\partial x_1\partial x_2} = x_2^{-1/2} ∂ x 1 2 ∂ 2 y = 2 , ∂ x 2 2 ∂ 2 y = − 2 1 x 1 x 2 − 3/2 , ∂ x 1 ∂ x 2 ∂ 2 y = x 2 − 1/2
Composite Functions and Multivariate Chain Rules ¶ In the univariate case, we learnt that the derivative of the composite function
y = f ( x ) = g ( h ( x ) ) y = f(x) = g(h(x)) y = f ( x ) = g ( h ( x )) is
d y d x = g ′ ( h ( x ) ) h ′ ( x ) . \frac{dy}{dx} = g'(h(x))h'(x). d x d y = g ′ ( h ( x )) h ′ ( x ) . A multivariate form of the chain rule can be used with multivariate compositefunctions.
Multivariate Chain Rule I (Single Parameter) ¶ If the arguments of the differentiable function
y = f ( x 1 , x 2 , … , x n ) y = f(x_1, x_2, \ldots, x_n) y = f ( x 1 , x 2 , … , x n ) are themselves differentiable functions of a single variable t t t
x 1 = g 1 ( t ) , x 2 = g 2 ( t ) , … , x n = g n ( t ) , x_1 = g^1(t), \quad x_2 = g^2(t), \quad \ldots, \quad x_n = g^n(t), x 1 = g 1 ( t ) , x 2 = g 2 ( t ) , … , x n = g n ( t ) , where g i ( t ) g^i(t) g i ( t ) i i i
d y d t = f 1 d x 1 d t + f 2 d x 2 d t + ⋯ + f n d x n d t , \frac{dy}{dt}
=
f_1 \frac{dx_1}{dt}
+
f_2 \frac{dx_2}{dt}
+
\cdots
+
f_n \frac{dx_n}{dt}, d t d y = f 1 d t d x 1 + f 2 d t d x 2 + ⋯ + f n d t d x n , where
f i = ∂ y ∂ x i . f_i = \frac{\partial y}{\partial x_i}. f i = ∂ x i ∂ y . Example
Consider the function
y = f ( x 1 , x 2 ) = x 1 2 x 2 , y = f(x_1, x_2) = x_1^2 x_2, y = f ( x 1 , x 2 ) = x 1 2 x 2 , where
x 1 = t 2 and x 2 = 3 t − 1. x_1 = t^2
\quad \text{and} \quad
x_2 = 3t - 1. x 1 = t 2 and x 2 = 3 t − 1. Then
d x 1 d t = 2 t and d x 2 d t = 3 , \frac{dx_1}{dt} = 2t
\quad \text{and} \quad
\frac{dx_2}{dt} = 3, d t d x 1 = 2 t and d t d x 2 = 3 , and
f 1 = 2 x 1 x 2 and f 2 = x 1 2 . f_1 = 2x_1 x_2
\quad \text{and} \quad
f_2 = x_1^2. f 1 = 2 x 1 x 2 and f 2 = x 1 2 . Using the chain rule,
d y d t = ( 2 x 1 x 2 ) d x 1 d t + ( x 1 2 ) d x 2 d t . \frac{dy}{dt}
=
(2x_1 x_2)\frac{dx_1}{dt}
+
(x_1^2)\frac{dx_2}{dt}. d t d y = ( 2 x 1 x 2 ) d t d x 1 + ( x 1 2 ) d t d x 2 . Substituting,
d y d t = ( 2 t 2 ( 3 t − 1 ) ) ( 2 t ) + 3 ( t 2 ) 2 = 4 t 3 ( 3 t − 1 ) + 3 t 4 . \frac{dy}{dt}
=
(2t^2(3t - 1))(2t)
+
3(t^2)^2
=
4t^3(3t - 1) + 3t^4. d t d y = ( 2 t 2 ( 3 t − 1 )) ( 2 t ) + 3 ( t 2 ) 2 = 4 t 3 ( 3 t − 1 ) + 3 t 4 . Simplifying,
d y d t = 15 t 4 − 4 t 3 . \frac{dy}{dt}
=
15t^4 - 4t^3. d t d y = 15 t 4 − 4 t 3 . Multivariate Chain Rule II (Multiple Parameters) ¶ If the arguments of the differentiable function
y = f ( x 1 , x 2 , … , x n ) y = f(x_1, x_2, \ldots, x_n) y = f ( x 1 , x 2 , … , x n ) are themselves differentiable functions of variables t 1 , … , t m t_1, \ldots, t_m t 1 , … , t m
x 1 = g 1 ( t 1 , … , t m ) , … , x n = g n ( t 1 , … , t m ) , x_1 = g^1(t_1, \ldots, t_m), \quad
\ldots, \quad
x_n = g^n(t_1, \ldots, t_m), x 1 = g 1 ( t 1 , … , t m ) , … , x n = g n ( t 1 , … , t m ) , then, for each t i t_i t i
∂ y ∂ t i = f 1 ∂ x 1 ∂ t i + f 2 ∂ x 2 ∂ t i + ⋯ + f n ∂ x n ∂ t i , \frac{\partial y}{\partial t_i}
=
f_1 \frac{\partial x_1}{\partial t_i}
+
f_2 \frac{\partial x_2}{\partial t_i}
+
\cdots
+
f_n \frac{\partial x_n}{\partial t_i}, ∂ t i ∂ y = f 1 ∂ t i ∂ x 1 + f 2 ∂ t i ∂ x 2 + ⋯ + f n ∂ t i ∂ x n , where
f i = ∂ y ∂ x i . f_i = \frac{\partial y}{\partial x_i}. f i = ∂ x i ∂ y . Partial Derivatives with Change of Variables ¶ For each of the following cases, find the partial derivatives ∂ Z ∂ u \displaystyle \frac{\partial Z}{\partial u} ∂ u ∂ Z ∂ Z ∂ v \displaystyle \frac{\partial Z}{\partial v} ∂ v ∂ Z
(i) Let Z = f ( x , y ) = 4 x 2 + 2 x y + y 2 , Z = f(x,y) = 4x^2 + 2xy + y^2, Z = f ( x , y ) = 4 x 2 + 2 x y + y 2 , x = 3 u 2 and y = u − 2 v x = 3u^2 \text{ and } y = u - 2v x = 3 u 2 and y = u − 2 v
(ii) Let Z = f ( x , y ) = a x 3 − b x 2 y + c y , Z = f(x,y) = ax^3 - bx^2y + cy, Z = f ( x , y ) = a x 3 − b x 2 y + cy , x = γ u + θ v and y = θ u − γ v 2 x = \gamma u + \theta v \text{ and } y = \theta u - \gamma v^2 x = γ u + θ v and y = θ u − γ v 2
(iii) Let Z = f ( x , y ) = 2 e x + 1 2 x 2 y − 4 y , Z = f(x,y) = 2e^x + \tfrac{1}{2}x^2y - 4y, Z = f ( x , y ) = 2 e x + 2 1 x 2 y − 4 y , x = 1 4 u and y = u 2 + 6 v x = \tfrac{1}{4}u \text{ and } y = u^2 + 6v x = 4 1 u and y = u 2 + 6 v
(iv) Let Z = f ( x , y , u ) = 2 x 3 − 3 x y 2 + 0.75 y u − 5 u 2 , Z = f(x,y,u) = 2x^3 - 3xy^2 + 0.75yu - 5u^2, Z = f ( x , y , u ) = 2 x 3 − 3 x y 2 + 0.75 y u − 5 u 2 , x = u + v and y = v 2 x = \sqrt{u+v} \text{ and } y = v^2 x = u + v and y = v 2
Answers
(i) ∂ Z ∂ u = 48 x u + 12 y u + 2 x + 2 y , and ∂ Z ∂ v = − 4 x − 4 y \frac{\partial Z}{\partial u} = 48xu + 12yu + 2x + 2y, \text{ and } \frac{\partial Z}{\partial v} = -4x - 4y ∂ u ∂ Z = 48 xu + 12 y u + 2 x + 2 y , and ∂ v ∂ Z = − 4 x − 4 y
(ii) ∂ Z ∂ u = ( 3 a x 2 − 2 b x y ) γ + ( − b x 2 + c ) θ , and ∂ Z ∂ v = ( 3 a x 2 − 2 b x y ) θ + ( − b x 2 + c ) ( − 2 γ v ) \frac{\partial Z}{\partial u} = (3ax^2 - 2bxy)\gamma + (-bx^2 + c)\theta, \text{ and } \frac{\partial Z}{\partial v} = (3ax^2 - 2bxy)\theta + (-bx^2 + c)(-2\gamma v) ∂ u ∂ Z = ( 3 a x 2 − 2 b x y ) γ + ( − b x 2 + c ) θ , and ∂ v ∂ Z = ( 3 a x 2 − 2 b x y ) θ + ( − b x 2 + c ) ( − 2 γ v )
(iii) ∂ Z ∂ u = ( 2 e x + x y ) 1 4 + ( 1 2 x 2 − 4 ) 2 u , and ∂ Z ∂ v = 3 x 2 − 24 \frac{\partial Z}{\partial u} = (2e^x + xy)\tfrac{1}{4} + \left(\tfrac{1}{2}x^2 - 4\right)2u, \text{ and } \frac{\partial Z}{\partial v} = 3x^2 - 24 ∂ u ∂ Z = ( 2 e x + x y ) 4 1 + ( 2 1 x 2 − 4 ) 2 u , and ∂ v ∂ Z = 3 x 2 − 24
(iv) ∂ Z ∂ u = ( 6 x 2 − 3 y 2 ) 1 2 u + v + 0.75 y − 10 u , and ∂ Z ∂ v = ( 6 x 2 − 3 y 2 ) 1 2 u + v + ( − 6 x y + 0.75 u ) 2 v \frac{\partial Z}{\partial u} = (6x^2 - 3y^2)\frac{1}{2\sqrt{u+v}} + 0.75y - 10u, \text{ and } \frac{\partial Z}{\partial v} = (6x^2 - 3y^2)\frac{1}{2\sqrt{u+v}} + (-6xy + 0.75u)2v ∂ u ∂ Z = ( 6 x 2 − 3 y 2 ) 2 u + v 1 + 0.75 y − 10 u , and ∂ v ∂ Z = ( 6 x 2 − 3 y 2 ) 2 u + v 1 + ( − 6 x y + 0.75 u ) 2 v
In parts (i)–(iii), the variable Z Z Z u u u v v v only through the intermediate variables x x x y y y only indirect effects .
In part (iv), however, Z = f ( x , y , u ) Z = f(x,y,u) Z = f ( x , y , u ) u u u both directly and indirectly (through x x x
∂ Z ∂ u = Z x ∂ x ∂ u + Z y ∂ y ∂ u + Z u . \frac{\partial Z}{\partial u}
=
Z_x \frac{\partial x}{\partial u}
+
Z_y \frac{\partial y}{\partial u}
+
Z_u. ∂ u ∂ Z = Z x ∂ u ∂ x + Z y ∂ u ∂ y + Z u . A very common mistake is to omit the direct term Z u Z_u Z u , which leads to an incomplete derivative.
Total Differentials ¶ The total differential of the multivariate function
y = f ( x 1 , … , x n ) y = f(x_1, \ldots, x_n) y = f ( x 1 , … , x n ) evaluated at the point
( x 1 0 , x 2 0 , … , x n 0 ) (x_1^0, x_2^0, \ldots, x_n^0) ( x 1 0 , x 2 0 , … , x n 0 ) is
d y = f 1 ( x 1 0 , … , x n 0 ) d x 1 + f 2 ( x 1 0 , … , x n 0 ) d x 2 + ⋯ + f n ( x 1 0 , … , x n 0 ) d x n . \begin{aligned}
dy
&=
f_1(x_1^0, \ldots, x_n^0)\,dx_1
+
f_2(x_1^0, \ldots, x_n^0)\,dx_2 \\
&\quad
+
\cdots
+
f_n(x_1^0, \ldots, x_n^0)\,dx_n.
\end{aligned} d y = f 1 ( x 1 0 , … , x n 0 ) d x 1 + f 2 ( x 1 0 , … , x n 0 ) d x 2 + ⋯ + f n ( x 1 0 , … , x n 0 ) d x n . where f i ( x 1 0 , x 2 0 , … , x n 0 ) f_i(x_1^0, x_2^0, \ldots, x_n^0) f i ( x 1 0 , x 2 0 , … , x n 0 ) f ( x 1 , x 2 , … , x n ) f(x_1, x_2, \ldots, x_n) f ( x 1 , x 2 , … , x n ) i i i ( x 1 0 , x 2 0 , … , x n 0 ) (x_1^0, x_2^0, \ldots, x_n^0) ( x 1 0 , x 2 0 , … , x n 0 )
Example
Given
z = f ( x , y ) = x 4 + 8 x y + 3 y 3 , z = f(x,y) = x^4 + 8xy + 3y^3, z = f ( x , y ) = x 4 + 8 x y + 3 y 3 , the total differential is
d z = z x d x + z y d y . dz = z_x\,dx + z_y\,dy. d z = z x d x + z y d y . Since
z x = 4 x 3 + 8 y and z y = 8 x + 9 y 2 , z_x = 4x^3 + 8y
\quad \text{and} \quad
z_y = 8x + 9y^2, z x = 4 x 3 + 8 y and z y = 8 x + 9 y 2 , we have
d z = ( 4 x 3 + 8 y ) d x + ( 8 x + 9 y 2 ) d y . dz = (4x^3 + 8y)\,dx + (8x + 9y^2)\,dy. d z = ( 4 x 3 + 8 y ) d x + ( 8 x + 9 y 2 ) d y . Higher-Order Total Differentials ¶ We can take higher-order total differentials if required. Continuing with the same example, the second-order total differential is
d 2 z = 12 x 2 d x 2 + 16 d x d y + 18 y d y 2 . d^2 z
=
12x^2\,dx^2
+
16\,dx\,dy
+
18y\,dy^2. d 2 z = 12 x 2 d x 2 + 16 d x d y + 18 y d y 2 . The total differential provides a direct link to the multivariate chain rule.
If each variable x i x_i x i t t t
d x i = d x i d t d t . dx_i = \frac{dx_i}{dt}\,dt. d x i = d t d x i d t . Substituting into the total differential,
d y = ∑ i = 1 n ∂ f ∂ x i d x i d t d t . dy
=
\sum_{i=1}^n
\frac{\partial f}{\partial x_i}
\frac{dx_i}{dt}\,dt. d y = i = 1 ∑ n ∂ x i ∂ f d t d x i d t . Dividing both sides by d t dt d t
d y d t = ∑ i = 1 n ∂ f ∂ x i d x i d t , \frac{dy}{dt}
=
\sum_{i=1}^n
\frac{\partial f}{\partial x_i}
\frac{dx_i}{dt}, d t d y = i = 1 ∑ n ∂ x i ∂ f d t d x i , which is precisely the multivariate chain rule .
Thus, the chain rule can be interpreted as applying the total differential to situations in which the arguments of a multivariate function depend on an underlying parameter.
Total Differentials ¶ 1. Find the total differentials of each function.
(i) w = 2 x 2 + 1 2 x y − 3 y 3 w = 2x^2 + \frac{1}{2}xy - 3y^3 w = 2 x 2 + 2 1 x y − 3 y 3 (ii) w = 4 x 1 3 − ln ( x 1 x 2 ) + 6 x 2 w = 4x_1^3 - \ln(x_1 x_2) + 6x_2 w = 4 x 1 3 − ln ( x 1 x 2 ) + 6 x 2 (iii) z = x 2 y 3 + x y z = \dfrac{x^2}{y^3 + xy} z = y 3 + x y x 2 (iv) y = 2 x 1 2 e 3 x 2 y = 2x_1^2 e^{3x_2} y = 2 x 1 2 e 3 x 2
2. For each function, use the total differential to approximate the change in y y y x x x z z z
(i) y = x 2 + 4 x − z 2 − 2 x z y = x^2 + 4x - z^2 - 2xz y = x 2 + 4 x − z 2 − 2 x z x = 1 x = 1 x = 1 z = 4 z = 4 z = 4 Δ x = 2 \Delta x = 2 Δ x = 2 Δ z = − 2 \Delta z = -2 Δ z = − 2
(ii) y = e x 2 + 3 z y = e^{x^2 + 3z} y = e x 2 + 3 z x = 1 x = 1 x = 1 z = 2 z = 2 z = 2 Δ x = 2 \Delta x = 2 Δ x = 2 Δ z = − 2 \Delta z = -2 Δ z = − 2
(iii) y = ln x 3 − 4 z + 2 x z y = \ln x^3 - 4z + 2xz y = ln x 3 − 4 z + 2 x z x = 1 x = 1 x = 1 z = 2 z = 2 z = 2 Δ x = 2 \Delta x = 2 Δ x = 2 Δ z = 4 \Delta z = 4 Δ z = 4
(iv) y = x 2 + e z / 2 y = x^2 + e^{z/2} y = x 2 + e z /2 x = 2 x = 2 x = 2 z = 2 z = 2 z = 2 Δ x = 1 \Delta x = 1 Δ x = 1 Δ z = − 1 \Delta z = -1 Δ z = − 1
Answers: Question 1
(i) d w = ( 4 x + 1 2 y ) , d x + ( 1 2 x − 9 y 2 ) , d y dw = (4x + \tfrac{1}{2}y),dx + (\tfrac{1}{2}x - 9y^2),dy d w = ( 4 x + 2 1 y ) , d x + ( 2 1 x − 9 y 2 ) , d y (ii) d w = ( 12 x 1 2 − 1 x 1 ) , d x 1 + ( 6 − 1 x 2 ) , d x 2 dw = \left(12x_1^2 - \frac{1}{x_1}\right),dx_1 + \left(6 - \frac{1}{x_2}\right),dx_2 d w = ( 12 x 1 2 − x 1 1 ) , d x 1 + ( 6 − x 2 1 ) , d x 2 (iii) d z = 2 x y 3 + x 2 y ( y 3 + x y ) 2 , d x − 3 x 2 y 2 + x 3 ( y 3 + x y ) 2 , d y dz = \dfrac{2xy^3 + x^2y}{(y^3 + xy)^2},dx - \dfrac{3x^2y^2 + x^3}{(y^3 + xy)^2},dy d z = ( y 3 + x y ) 2 2 x y 3 + x 2 y , d x − ( y 3 + x y ) 2 3 x 2 y 2 + x 3 , d y (iv) d y = ( 4 x 1 e 3 x 2 ) , d x 1 + ( 6 x 1 2 e 3 x 2 ) , d x 2 dy = (4x_1 e^{3x_2}),dx_1 + (6x_1^2 e^{3x_2}),dx_2 d y = ( 4 x 1 e 3 x 2 ) , d x 1 + ( 6 x 1 2 e 3 x 2 ) , d x 2
Answers: Question 2
(i) Δ y = 24 \Delta y = 24 Δ y = 24 d y = 16 dy = 16 d y = 16 (ii) Δ y = 7007 \Delta y = 7007 Δ y = 7007 d y = − 2194 dy = -2194 d y = − 2194 (iii) Δ y = 19.3 \Delta y = 19.3 Δ y = 19.3 d y = 6 dy = 6 d y = 6 (iv) d y = 2 x , d x + 1 2 e z / 2 , d z = 4 − 1 2 e = 2.64 dy = 2x,dx + \tfrac{1}{2}e^{z/2},dz = 4 - \tfrac{1}{2}e = 2.64 d y = 2 x , d x + 2 1 e z /2 , d z = 4 − 2 1 e = 2.64 Δ y = 3.93 \Delta y = 3.93 Δ y = 3.93
The total differential provides a local linear approximation to the actual change in a function.
When changes are small, d y dy d y Δ y \Delta y Δ y
Total Derivatives ¶ Given a case with
z = f ( x , y ) z = f(x,y) z = f ( x , y ) y = g ( x ) y = g(x) y = g ( x ) x x x y y y x x x z z z f f f g g g
The total derivative measures the direct effect of x x x z z z ∂ z / ∂ x \partial z / \partial x ∂ z / ∂ x indirect effect of x x x z z z y y y ( ∂ z / ∂ y ) ( d y / d x ) (\partial z/\partial y)(dy/dx) ( ∂ z / ∂ y ) ( d y / d x )
That is,
d z d x = z x + z y d y d x . \frac{dz}{dx}
=
z_x + z_y \frac{dy}{dx}. d x d z = z x + z y d x d y . An alternative method of finding the total derivative is to take the total differential of z z z
d z = z x , d x + z y , d y . dz = z_x,dx + z_y,dy. d z = z x , d x + z y , d y . Dividing through by d x dx d x
d z d x = z x d x d x + z y d y d x . \frac{dz}{dx}
=
z_x \frac{dx}{dx}
+
z_y \frac{dy}{dx}. d x d z = z x d x d x + z y d x d y . Since d x / d x = 1 dx/dx = 1 d x / d x = 1
d z d x = z x + z y d y d x . \frac{dz}{dx}
=
z_x + z_y \frac{dy}{dx}. d x d z = z x + z y d x d y . Example 1
Let
z = f ( x , y ) = 6 x 3 + 7 y , z = f(x,y) = 6x^3 + 7y, z = f ( x , y ) = 6 x 3 + 7 y , where
y = 4 x 2 + 3 x . y = 4x^2 + 3x. y = 4 x 2 + 3 x . Then
z x = 18 x 2 , z y = 7 , d y d x = 8 x + 3. z_x = 18x^2,
\qquad
z_y = 7,
\qquad
\frac{dy}{dx} = 8x + 3. z x = 18 x 2 , z y = 7 , d x d y = 8 x + 3. Substituting into the total derivative formula,
d z d x = 18 x 2 + 7 ( 8 x + 3 ) = 18 x 2 + 56 x + 21. \frac{dz}{dx}
=
18x^2 + 7(8x + 3)
=
18x^2 + 56x + 21. d x d z = 18 x 2 + 7 ( 8 x + 3 ) = 18 x 2 + 56 x + 21. Example 2 (Parametric Dependence)
Let
z = f ( x , y ) = 8 x 2 + 3 y 2 , z = f(x,y) = 8x^2 + 3y^2, z = f ( x , y ) = 8 x 2 + 3 y 2 , where
x = 4 t , y = 5 t . x = 4t,
\qquad
y = 5t. x = 4 t , y = 5 t . The total derivative of z z z t t t
d z d t = z x d x d t + z y d y d t . \frac{dz}{dt}
=
z_x \frac{dx}{dt}
+
z_y \frac{dy}{dt}. d t d z = z x d t d x + z y d t d y . Since
z x = 16 x , z y = 6 y , d x d t = 4 , d y d t = 5 , z_x = 16x,
\qquad
z_y = 6y,
\qquad
\frac{dx}{dt} = 4,
\qquad
\frac{dy}{dt} = 5, z x = 16 x , z y = 6 y , d t d x = 4 , d t d y = 5 , we obtain
d z d t = 16 x ( 4 ) + 6 y ( 5 ) = 64 x + 30 y . \frac{dz}{dt}
=
16x(4) + 6y(5)
=
64x + 30y. d t d z = 16 x ( 4 ) + 6 y ( 5 ) = 64 x + 30 y . Substituting x = 4 t x = 4t x = 4 t y = 5 t y = 5t y = 5 t
d z d t = 64 ( 4 t ) + 30 ( 5 t ) = 256 t + 150 t = 406 t . \frac{dz}{dt}
=
64(4t) + 30(5t)
=
256t + 150t
=
406t. d t d z = 64 ( 4 t ) + 30 ( 5 t ) = 256 t + 150 t = 406 t . Total Derivatives ¶ In the following questions, the function f ( x , y ) f(x,y) f ( x , y ) x x x both directly and indirectly through y y y total derivative , not a partial derivative.
Recall that if y = y ( t ) y = y(t) y = y ( t )
d f d t = f x d x d t + f y d y d t . \frac{df}{dt} = f_x \frac{dx}{dt} + f_y \frac{dy}{dt}. d t df = f x d t d x + f y d t d y . 1. Total derivative with respect to x x x
Find the total derivative d f ( x , y ) / d x df(x,y)/dx df ( x , y ) / d x
(i) f ( x , y ) = 6 x 2 + 15 x y + 3 y 2 , where y = 7 x 2 . f(x,y) = 6x^2 + 15xy + 3y^2,
\quad \text{where } y = 7x^2. f ( x , y ) = 6 x 2 + 15 x y + 3 y 2 , where y = 7 x 2 .
(ii) f ( x , y ) = 9 x − 7 y 2 x + 5 y , where y = 3 x − 4. f(x,y) = \frac{9x - 7y}{2x + 5y},
\quad \text{where } y = 3x - 4. f ( x , y ) = 2 x + 5 y 9 x − 7 y , where y = 3 x − 4.
(iii) f ( x , y ) = 8 x − 12 y , where y = x + 1 x 2 . f(x,y) = 8x - 12y,
\quad \text{where } y = \frac{x+1}{x^2}. f ( x , y ) = 8 x − 12 y , where y = x 2 x + 1 .
2. Total derivative with respect to w w w
Find the total derivative d f ( x , y ) / d w df(x,y)/dw df ( x , y ) / d w
(i) f ( x , y ) = 7 x 2 + 4 y 2 , where x = 5 w and y = 4 w . f(x,y) = 7x^2 + 4y^2,
\quad \text{where } x = 5w \text{ and } y = 4w. f ( x , y ) = 7 x 2 + 4 y 2 , where x = 5 w and y = 4 w .
(ii) f ( x , y ) = 10 x 2 − 6 x y − 12 y 2 , where x = 2 w and y = 3 w . f(x,y) = 10x^2 - 6xy - 12y^2,
\quad \text{where } x = 2w \text{ and } y = 3w. f ( x , y ) = 10 x 2 − 6 x y − 12 y 2 , where x = 2 w and y = 3 w .
Do not compute f x f_x f x
Because y y y x x x x x x w w w both terms :
d f d t = f x d x d t + f y d y d t . \frac{df}{dt} = f_x \frac{dx}{dt} + f_y \frac{dy}{dt}. d t df = f x d t d x + f y d t d y . Answers: Question 1
(i) d f d x = f x + f y d y d x = 210 x 2 + 84 x y + 12 x + 15 y . \frac{df}{dx}
= f_x + f_y \frac{dy}{dx}
= 210x^2 + 84xy + 12x + 15y. d x df = f x + f y d x d y = 210 x 2 + 84 x y + 12 x + 15 y .
(ii) d f d x = 59 ( y − 3 x ) ( 2 x + 5 y ) 2 . \frac{df}{dx}
= \frac{59(y - 3x)}{(2x + 5y)^2}. d x df = ( 2 x + 5 y ) 2 59 ( y − 3 x ) .
(iii) d f d x = 8 + 12 ( x + 2 ) x 3 . \frac{df}{dx}
= 8 + \frac{12(x+2)}{x^3}. d x df = 8 + x 3 12 ( x + 2 ) .
Answers: Question 2
(i) d f d w = f x d x d w + f y d y d w = 14 x ( 5 ) + 8 y ( 4 ) = 70 x + 32 y . \frac{df}{dw}
= f_x \frac{dx}{dw} + f_y \frac{dy}{dw}
= 14x(5) + 8y(4)
= 70x + 32y. d w df = f x d w d x + f y d w d y = 14 x ( 5 ) + 8 y ( 4 ) = 70 x + 32 y .
(ii) d f d w = ( 20 x − 6 y ) ( 2 ) + ( − 6 x − 24 y ) ( 3 ) = 22 x − 84 y . \frac{df}{dw}
= (20x - 6y)(2) + (-6x - 24y)(3)
= 22x - 84y. d w df = ( 20 x − 6 y ) ( 2 ) + ( − 6 x − 24 y ) ( 3 ) = 22 x − 84 y .
Implicit Multivariate Differentiation ¶ An explicit function expresses the dependent variable directly as a function of independent variables:
y = f ( x 1 , x 2 , … , x n ) . y = f(x_1, x_2, \ldots, x_n). y = f ( x 1 , x 2 , … , x n ) . An implicit function combines the dependent and independent variables in a relation of the form
F ( y , x 1 , x 2 , … , x n ) = k , F(y, x_1, x_2, \ldots, x_n) = k, F ( y , x 1 , x 2 , … , x n ) = k , where k k k
Implicit Function Rule
For an implicit function
F ( y , x 1 , x 2 , … , x n ) = k , F(y, x_1, x_2, \ldots, x_n) = k, F ( y , x 1 , x 2 , … , x n ) = k , defined at a point ( y 0 , x 1 0 , x 2 0 , … , x n 0 ) (y^0, x_1^0, x_2^0, \ldots, x_n^0) ( y 0 , x 1 0 , x 2 0 , … , x n 0 ) F F F
F y ( y 0 , x 1 0 , x 2 0 , … , x n 0 ) ≠ 0. F_y(y^0, x_1^0, x_2^0, \ldots, x_n^0) \neq 0. F y ( y 0 , x 1 0 , x 2 0 , … , x n 0 ) = 0. Then there exists a function
y = f ( x 1 , x 2 , … , x n ) y = f(x_1, x_2, \ldots, x_n) y = f ( x 1 , x 2 , … , x n ) defined in a neighborhood of ( x 1 0 , x 2 0 , … , x n 0 ) (x_1^0, x_2^0, \ldots, x_n^0) ( x 1 0 , x 2 0 , … , x n 0 )
F ( f ( x 1 0 , x 2 0 , … , x n 0 ) , x 1 0 , x 2 0 , … , x n 0 ) = k , F(f(x_1^0, x_2^0, \ldots, x_n^0), x_1^0, x_2^0, \ldots, x_n^0) = k, F ( f ( x 1 0 , x 2 0 , … , x n 0 ) , x 1 0 , x 2 0 , … , x n 0 ) = k , y 0 = f ( x 1 0 , x 2 0 , … , x n 0 ) , y^0 = f(x_1^0, x_2^0, \ldots, x_n^0), y 0 = f ( x 1 0 , x 2 0 , … , x n 0 ) , and
f i ( x 1 0 , x 2 0 , … , x n 0 ) = − F x i ( y 0 , x 1 0 , x 2 0 , … , x n 0 ) F y ( y 0 , x 1 0 , x 2 0 , … , x n 0 ) . f_i(x_1^0, x_2^0, \ldots, x_n^0)
=
-
\frac{
F_{x_i}(y^0, x_1^0, x_2^0, \ldots, x_n^0)
}{
F_y(y^0, x_1^0, x_2^0, \ldots, x_n^0)
}. f i ( x 1 0 , x 2 0 , … , x n 0 ) = − F y ( y 0 , x 1 0 , x 2 0 , … , x n 0 ) F x i ( y 0 , x 1 0 , x 2 0 , … , x n 0 ) . Here,
F x i = ∂ F ∂ x i , F y = ∂ F ∂ y , F_{x_i} = \frac{\partial F}{\partial x_i},
\qquad
F_y = \frac{\partial F}{\partial y}, F x i = ∂ x i ∂ F , F y = ∂ y ∂ F , evaluated at ( y 0 , x 1 0 , x 2 0 , … , x n 0 ) (y^0, x_1^0, x_2^0, \ldots, x_n^0) ( y 0 , x 1 0 , x 2 0 , … , x n 0 )
This result is sometimes referred to as the inverse rule for implicit functions .
Implicit and Inverse Function Rule ¶ Find the derivative of each implicit function, where
d y d x = − F x F y , \frac{dy}{dx} = -\frac{F_x}{F_y}, d x d y = − F y F x , provided that F y ≠ 0 F_y \neq 0 F y = 0
(i) F ( x , y ) = x 2 + y 2 + ( x y ) 1 / 3 = 0 F(x,y) = x^2 + y^2 + (xy)^{1/3} = 0 F ( x , y ) = x 2 + y 2 + ( x y ) 1/3 = 0
(ii) F ( x , y ) = x 2 y + y 2 x + x y = 0 F(x,y) = x^2 y + y^2 x + xy = 0 F ( x , y ) = x 2 y + y 2 x + x y = 0
(iii) F ( x , y ) = ln x 3 + ( x y ) 2 − 4 y = 0 F(x,y) = \ln x^3 + (xy)^2 - 4y = 0 F ( x , y ) = ln x 3 + ( x y ) 2 − 4 y = 0
(iv) F ( x , y , w ) = w 3 y 3 + x 2 + w x y + 7 = 0 F(x,y,w) = w^3 y^3 + x^2 + wxy + 7 = 0 F ( x , y , w ) = w 3 y 3 + x 2 + w x y + 7 = 0 ∂ y / ∂ x \partial y / \partial x ∂ y / ∂ x
(v) F ( x , y ) = x y 2 e y = 0 F(x,y) = xy^2 e^y = 0 F ( x , y ) = x y 2 e y = 0
Answers
(i) d y d x = − 6 x + x − 2 / 3 y 1 / 3 6 y + y − 2 / 3 x 1 / 3 \frac{dy}{dx} = -\frac{6x + x^{-2/3} y^{1/3}}{6y + y^{-2/3} x^{1/3}} d x d y = − 6 y + y − 2/3 x 1/3 6 x + x − 2/3 y 1/3 (ii) d y d x = − y ( 2 x + y + 1 ) x ( 2 y + x + 1 ) \frac{dy}{dx} = -\frac{y(2x + y + 1)}{x(2y + x + 1)} d x d y = − x ( 2 y + x + 1 ) y ( 2 x + y + 1 ) (iii) d y d x = − 3 x − 1 + 2 x y 2 2 x 2 y − 4 \frac{dy}{dx} = -\frac{3x^{-1} + 2xy^2}{2x^2 y - 4} d x d y = − 2 x 2 y − 4 3 x − 1 + 2 x y 2 (iv) ∂ y ∂ x = − 3 x 2 + w y 3 w 3 y 2 + w x \frac{\partial y}{\partial x} = -\frac{3x^2 + wy}{3w^3 y^2 + wx} ∂ x ∂ y = − 3 w 3 y 2 + w x 3 x 2 + w y (v) d y d x = − y x ( 2 + y ) \frac{dy}{dx} = -\frac{y}{x(2 + y)} d x d y = − x ( 2 + y ) y
Homogeneous Functions and Euler’s Theorem ¶ A function y = f ( x 1 , … , x n ) y = f(x_1, \ldots, x_n) y = f ( x 1 , … , x n ) homogeneous of degree k k k if, for any s > 0 s > 0 s > 0
f ( s x 1 , … , s x n ) = s k f ( x 1 , … , x n ) . f(sx_1, \ldots, sx_n) = s^k f(x_1, \ldots, x_n). f ( s x 1 , … , s x n ) = s k f ( x 1 , … , x n ) . To check homogeneity, scale all arguments by a common factor s s s f ( s x , s y , s w ) = s k f ( x , y , w ) f(sx, sy, sw) = s^k f(x,y,w) f ( s x , sy , s w ) = s k f ( x , y , w ) k k k
If different terms scale with different powers of s s s , the function is not homogeneous .
Homogeneous Functions ¶ Which of the following functions are homogeneous? If homogeneous, what are their degrees?
1. f ( x , y ) = x 2 y + x 4 x 2 + y 2 f(x,y) = x^2y + \frac{x^4}{\sqrt{x^2 + y^2}} f ( x , y ) = x 2 y + x 2 + y 2 x 4 2. f ( x , y ) = x y + 5 + e x 2 / y 2 f(x,y) = \frac{x}{y} + 5 + e^{x^2/y^2} f ( x , y ) = y x + 5 + e x 2 / y 2 3. f ( x , y ) = x 2 + x sin y f(x,y) = x^2 + x \sin y f ( x , y ) = x 2 + x sin y
Solutions
If f ( x , y ) = x 2 y + x 4 x 2 + y 2 f(x, y) = x^2y + \frac{x^4}{\sqrt{x^2 + y^2}} f ( x , y ) = x 2 y + x 2 + y 2 x 4 ( c x , c y ) (cx, cy) ( c x , cy )
f ( c x , c y ) = ( c x ) 2 ( c y ) + ( c x ) 4 ( c x ) 2 + ( c y ) 2 = c 3 x 2 y + c 4 x 4 c x 2 + y 2 = c 3 x 2 y + c 3 x 4 x 2 + y 2 \begin{aligned}
f(cx, cy) &= (cx)^2(cy) + \frac{(cx)^4}{\sqrt{(cx)^2 + (cy)^2}} \\
&= c^3x^2y + \frac{c^4x^4}{c\sqrt{x^2 + y^2}} \\
&= c^3x^2y + c^3 \frac{x^4}{\sqrt{x^2 + y^2}}
\end{aligned} f ( c x , cy ) = ( c x ) 2 ( cy ) + ( c x ) 2 + ( cy ) 2 ( c x ) 4 = c 3 x 2 y + c x 2 + y 2 c 4 x 4 = c 3 x 2 y + c 3 x 2 + y 2 x 4 Which is c 3 f ( x , y ) c^3f(x, y) c 3 f ( x , y ) homogeneous of degree 3 .
If f ( x , y ) = x y + 5 + e x 2 / y 2 f(x, y) = \frac{x}{y} + 5 + e^{x^2/y^2} f ( x , y ) = y x + 5 + e x 2 / y 2
f ( c x , c y ) = c x c y + 5 + e c 2 x 2 / ( c 2 y 2 ) = x y + 5 + e x 2 / y 2 f(cx, cy) = \frac{cx}{cy} + 5 + e^{c^2x^2/(c^2y^2)} = \frac{x}{y} + 5 + e^{x^2/y^2} f ( c x , cy ) = cy c x + 5 + e c 2 x 2 / ( c 2 y 2 ) = y x + 5 + e x 2 / y 2 In other words, f ( c x , c y ) = f ( x , y ) f(cx, cy) = f(x, y) f ( c x , cy ) = f ( x , y ) f f f homogeneous of degree 0 (since f ( c x , c y ) = c 0 f ( x , y ) f(cx, cy) = c^0f(x, y) f ( c x , cy ) = c 0 f ( x , y )
For f ( x , y ) = x 2 + x sin y f(x, y) = x^2 + x \sin y f ( x , y ) = x 2 + x sin y
f ( c x , c y ) = c 2 x 2 + c x sin ( c y ) f(cx, cy) = c^2x^2 + cx \sin(cy) f ( c x , cy ) = c 2 x 2 + c x sin ( cy ) This cannot be written as c D f ( x , y ) c^D f(x, y) c D f ( x , y ) D D D x 2 x^2 x 2 D D D sin ( c y ) = c sin y \sin(cy) = c \sin y sin ( cy ) = c sin y y y y
So the function is not homogeneous.
Determine whether each function is homogeneous and, if so, state its degree of homogeneity . ¶ f ( x , y , w ) = x w + 3 y 5 x f(x,y,w) = \dfrac{x}{w} + \dfrac{3y}{5x} f ( x , y , w ) = w x + 5 x 3 y
f ( x , y , w ) = x 2 w + 2 w 2 y f(x,y,w) = \dfrac{x^2}{w} + \dfrac{2w^2}{y} f ( x , y , w ) = w x 2 + y 2 w 2
f ( x , y , w ) = x 3 y w + 2 x y w f(x,y,w) = \dfrac{x^3 y}{w} + 2xyw f ( x , y , w ) = w x 3 y + 2 x y w
f ( x , y ) = x 2 + y 2 f(x,y) = \sqrt{x^2 + y^2} f ( x , y ) = x 2 + y 2
f ( x , y , w ) = 3 x 2 y − 3 y w 2 f(x,y,w) = 3x^2 y - \dfrac{3y}{w^2} f ( x , y , w ) = 3 x 2 y − w 2 3 y
f ( x , y ) = x 1 / 2 y 1 / 4 + y 5 / 8 f(x,y) = x^{1/2} y^{1/4} + y^{5/8} f ( x , y ) = x 1/2 y 1/4 + y 5/8
Answers
1. Homogeneous of degree 0 2. Homogeneous of degree 1 3. Homogeneous of degree 3 4. Homogeneous of degree 1 5. Not homogeneous 6. Not homogeneous
Euler’s Theorem ¶ If y = f ( x 1 , … , x n ) y = f(x_1, \ldots, x_n) y = f ( x 1 , … , x n ) k k k
k y = x 1 f 1 ( x 1 , … , x n ) + ⋯ + x n f n ( x 1 , … , x n ) , k y
=
x_1 f_1(x_1, \ldots, x_n)
+
\cdots
+
x_n f_n(x_1, \ldots, x_n), k y = x 1 f 1 ( x 1 , … , x n ) + ⋯ + x n f n ( x 1 , … , x n ) , where
f i = ∂ f ∂ x i . f_i = \frac{\partial f}{\partial x_i}. f i = ∂ x i ∂ f . Proof of Euler’s Theorem ¶ By homogeneity,
f ( s x 1 , … , s x n ) = s k y . f(sx_1, \ldots, sx_n) = s^k y. f ( s x 1 , … , s x n ) = s k y . Taking the derivative with respect to s s s
d d s f ( s x 1 , … , s x n ) = x 1 f 1 ( s x 1 , … , s x n ) + ⋯ + x n f n ( s x 1 , … , s x n ) . \frac{d}{ds} f(sx_1, \ldots, sx_n)
=
x_1 f_1(sx_1, \ldots, sx_n)
+
\cdots
+
x_n f_n(sx_1, \ldots, sx_n). d s d f ( s x 1 , … , s x n ) = x 1 f 1 ( s x 1 , … , s x n ) + ⋯ + x n f n ( s x 1 , … , s x n ) . Differentiating the right-hand side,
d d s ( s k y ) = k s k − 1 y . \frac{d}{ds}(s^k y) = k s^{k-1} y. d s d ( s k y ) = k s k − 1 y . Setting s = 1 s = 1 s = 1
Notes
Note 1.
Any function homogeneous of degree 0 can be written as
f ! ( x 1 x i , x 2 x i , … , 1 , … , x n x i ) , f!\left(
\frac{x_1}{x_i}, \frac{x_2}{x_i}, \ldots, 1, \ldots, \frac{x_n}{x_i}
\right), f ! ( x i x 1 , x i x 2 , … , 1 , … , x i x n ) , for any i = 1 , … , n i = 1, \ldots, n i = 1 , … , n
Note 2.
If f ( x 1 , … , x n ) f(x_1, \ldots, x_n) f ( x 1 , … , x n ) k k k
f i = ∂ f ∂ x i f_i = \frac{\partial f}{\partial x_i} f i = ∂ x i ∂ f is homogeneous of degree k − 1 k-1 k − 1
Homogeneous Production Function and Euler’s Theorem ¶ Consider the production function
f ( x 1 , x 2 ) = x 1 1 / 4 x 2 1 / 3 . f(x_1,x_2) = x_1^{1/4} x_2^{1/3}. f ( x 1 , x 2 ) = x 1 1/4 x 2 1/3 . Question
i. Determine whether this production function is homogeneous. If so, of what degree?
ii. Take the partial derivatives of this production function and show that they are homogeneous of degree k − 1 k-1 k − 1
iii. Using Euler’s Theorem, show that
x 1 f 1 ( s x 1 , s x 2 ) + x 2 f 2 ( s x 1 , s x 2 ) = k s k − 1 f ( x 1 , x 2 ) . x_1 f_1(sx_1,sx_2) + x_2 f_2(sx_1,sx_2)
=
k s^{k-1} f(x_1,x_2). x 1 f 1 ( s x 1 , s x 2 ) + x 2 f 2 ( s x 1 , s x 2 ) = k s k − 1 f ( x 1 , x 2 ) . Answers
i. The function is homogeneous of degree
k = 1 4 + 1 3 = 7 12 , k = \frac14 + \frac13 = \frac{7}{12}, k = 4 1 + 3 1 = 12 7 , since
f ( s x 1 , s x 2 ) = ( s x 1 ) 1 / 4 ( s x 2 ) 1 / 3 = s 7 / 12 f ( x 1 , x 2 ) . f(sx_1,sx_2)
=
(sx_1)^{1/4}(sx_2)^{1/3}
=
s^{7/12} f(x_1,x_2). f ( s x 1 , s x 2 ) = ( s x 1 ) 1/4 ( s x 2 ) 1/3 = s 7/12 f ( x 1 , x 2 ) . ii. The partial derivatives are
f 1 ( x 1 , x 2 ) = ∂ f ∂ x 1 = 1 4 x 1 − 3 / 4 x 2 1 / 3 , f_1(x_1,x_2)
=
\frac{\partial f}{\partial x_1}
=
\frac14 x_1^{-3/4} x_2^{1/3}, f 1 ( x 1 , x 2 ) = ∂ x 1 ∂ f = 4 1 x 1 − 3/4 x 2 1/3 , and
f 2 ( x 1 , x 2 ) = ∂ f ∂ x 2 = 1 3 x 1 1 / 4 x 2 − 2 / 3 . f_2(x_1,x_2)
=
\frac{\partial f}{\partial x_2}
=
\frac13 x_1^{1/4} x_2^{-2/3}. f 2 ( x 1 , x 2 ) = ∂ x 2 ∂ f = 3 1 x 1 1/4 x 2 − 2/3 . Each partial derivative is homogeneous of degree
k − 1 = 7 12 − 1 = − 5 12 , k-1 = \frac{7}{12} - 1 = -\frac{5}{12}, k − 1 = 12 7 − 1 = − 12 5 , since
f 1 ( s x 1 , s x 2 ) = s − 5 / 12 f 1 ( x 1 , x 2 ) , f 2 ( s x 1 , s x 2 ) = s − 5 / 12 f 2 ( x 1 , x 2 ) . f_1(sx_1,sx_2) = s^{-5/12} f_1(x_1,x_2),
\quad
f_2(sx_1,sx_2) = s^{-5/12} f_2(x_1,x_2). f 1 ( s x 1 , s x 2 ) = s − 5/12 f 1 ( x 1 , x 2 ) , f 2 ( s x 1 , s x 2 ) = s − 5/12 f 2 ( x 1 , x 2 ) . iii. By Euler’s Theorem for homogeneous functions,
x 1 f 1 ( x 1 , x 2 ) + x 2 f 2 ( x 1 , x 2 ) = k f ( x 1 , x 2 ) . x_1 f_1(x_1,x_2) + x_2 f_2(x_1,x_2)
=
k f(x_1,x_2). x 1 f 1 ( x 1 , x 2 ) + x 2 f 2 ( x 1 , x 2 ) = k f ( x 1 , x 2 ) . Evaluating this expression at ( s x 1 , s x 2 ) (sx_1,sx_2) ( s x 1 , s x 2 )
x 1 f 1 ( s x 1 , s x 2 ) + x 2 f 2 ( s x 1 , s x 2 ) = k s k − 1 f ( x 1 , x 2 ) , x_1 f_1(sx_1,sx_2) + x_2 f_2(sx_1,sx_2)
=
k s^{k-1} f(x_1,x_2), x 1 f 1 ( s x 1 , s x 2 ) + x 2 f 2 ( s x 1 , s x 2 ) = k s k − 1 f ( x 1 , x 2 ) , where k = 7 12 k = \tfrac{7}{12} k = 12 7 f ( x 1 , x 2 ) = x 1 1 / 4 x 2 1 / 3 f(x_1,x_2) = x_1^{1/4} x_2^{1/3} f ( x 1 , x 2 ) = x 1 1/4 x 2 1/3
Homothetic Functions ¶ A homothetic function is a monotonic transformation of a homogeneous function.
That is, if
y = f ( x 1 , … , x n ) y = f(x_1, \ldots, x_n) y = f ( x 1 , … , x n ) is a homogeneous function, then
is a homothetic function if g ( y ) g(y) g ( y )
g ′ ( y ) > 0 for all y or g ′ ( y ) < 0 for all y . g'(y) > 0 \quad \text{for all } y
\quad \text{or} \quad
g'(y) < 0 \quad \text{for all } y. g ′ ( y ) > 0 for all y or g ′ ( y ) < 0 for all y . Note. While every homogeneous function is a homothetic function (since we can simply choose g ( y ) = y g(y) = y g ( y ) = y
Example
For example, take
y = x 1 α x 2 β , y = x_1^{\alpha} x_2^{\beta}, y = x 1 α x 2 β , which is homogeneous of degree α + β \alpha + \beta α + β
Taking logarithms, we obtain
z = ln ( y ) = α ln ( x 1 ) + β ln ( x 2 ) . z = \ln(y) = \alpha \ln(x_1) + \beta \ln(x_2). z = ln ( y ) = α ln ( x 1 ) + β ln ( x 2 ) . The function z z z
However, this homothetic function is not homogeneous in the arguments x 1 x_1 x 1 x 2 x_2 x 2
α ln ( s x 1 ) + β ln ( s x 2 ) = α ln ( x 1 ) + β ln ( x 2 ) + ( α + β ) ln ( s ) \alpha \ln(sx_1) + \beta \ln(sx_2)
=
\alpha \ln(x_1) + \beta \ln(x_2) + (\alpha + \beta)\ln(s) α ln ( s x 1 ) + β ln ( s x 2 ) = α ln ( x 1 ) + β ln ( x 2 ) + ( α + β ) ln ( s ) and therefore
z ( s x 1 , s x 2 ) = z ( x 1 , x 2 ) + ( α + β ) ln ( s ) , z(sx_1, sx_2)
=
z(x_1, x_2) + (\alpha + \beta)\ln(s), z ( s x 1 , s x 2 ) = z ( x 1 , x 2 ) + ( α + β ) ln ( s ) , which does not satisfy the definition of homogeneity.
Homogeneous and Homothetic Functions ¶ Consider the function
y = f ( x 1 , x 2 ) = x 1 x 2 y = f(x_1, x_2) = x_1 x_2 y = f ( x 1 , x 2 ) = x 1 x 2 defined over the domain x 1 > 0 x_1 > 0 x 1 > 0 x 2 > 0 x_2 > 0 x 2 > 0
g ( y ) = ln y , h ( y ) = 10 y , j ( y ) = y 2 , k ( y ) = e y . g(y) = \ln y,
\quad
h(y) = 10y,
\quad
j(y) = y^2,
\quad
k(y) = e^y. g ( y ) = ln y , h ( y ) = 10 y , j ( y ) = y 2 , k ( y ) = e y . Answer the following questions.
Is f ( x 1 , x 2 ) f(x_1, x_2) f ( x 1 , x 2 )
Is g ( y ) g(y) g ( y ) g ( y ) g(y) g ( y ) x 1 x_1 x 1 x 2 x_2 x 2
How about h ( y ) h(y) h ( y ) x 1 x_1 x 1 x 2 x_2 x 2
How about j ( y ) j(y) j ( y ) k ( y ) k(y) k ( y ) ( x 1 , x 2 ) (x_1, x_2) ( x 1 , x 2 )
Answers
f ( x 1 , x 2 ) f(x_1, x_2) f ( x 1 , x 2 )
Yes. Since
f ( s x 1 , s x 2 ) = ( s x 1 ) ( s x 2 ) = s 2 f ( x 1 , x 2 ) , f(sx_1, sx_2) = (sx_1)(sx_2) = s^2 f(x_1, x_2), f ( s x 1 , s x 2 ) = ( s x 1 ) ( s x 2 ) = s 2 f ( x 1 , x 2 ) , the function is homogeneous of degree 2 .
g ( y ) = ln y g(y)=\ln y g ( y ) = ln y
Homothetic? Yes, because ln ( ⋅ ) \ln(\cdot) ln ( ⋅ ) ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) y = x 1 x 2 > 0 y=x_1x_2>0 y = x 1 x 2 > 0
Homogeneous in ( x 1 , x 2 ) (x_1,x_2) ( x 1 , x 2 ) No.G ( x 1 , x 2 ) = ln ( f ( x 1 , x 2 ) ) = ln ( x 1 x 2 ) G(x_1,x_2)=\ln(f(x_1,x_2))=\ln(x_1x_2) G ( x 1 , x 2 ) = ln ( f ( x 1 , x 2 )) = ln ( x 1 x 2 )
G ( s x 1 , s x 2 ) = ln ( ( s x 1 ) ( s x 2 ) ) = ln ( s 2 x 1 x 2 ) = ln ( x 1 x 2 ) + 2 ln s , G(sx_1,sx_2)=\ln\big((sx_1)(sx_2)\big)=\ln(s^2x_1x_2)=\ln(x_1x_2)+2\ln s, G ( s x 1 , s x 2 ) = ln ( ( s x 1 ) ( s x 2 ) ) = ln ( s 2 x 1 x 2 ) = ln ( x 1 x 2 ) + 2 ln s , which is not of the form s k G ( x 1 , x 2 ) s^k G(x_1,x_2) s k G ( x 1 , x 2 ) k k k
h ( y ) = 10 y h(y)=10y h ( y ) = 10 y
Homothetic? Yes (it is strictly increasing in y y y
Homogeneous in ( x 1 , x 2 ) (x_1,x_2) ( x 1 , x 2 ) Yes.H ( x 1 , x 2 ) = h ( f ( x 1 , x 2 ) ) = 10 x 1 x 2 H(x_1,x_2)=h(f(x_1,x_2))=10x_1x_2 H ( x 1 , x 2 ) = h ( f ( x 1 , x 2 )) = 10 x 1 x 2
H ( s x 1 , s x 2 ) = 10 ( s x 1 ) ( s x 2 ) = s 2 ⋅ 10 x 1 x 2 = s 2 H ( x 1 , x 2 ) , H(sx_1,sx_2)=10(sx_1)(sx_2)=s^2\cdot 10x_1x_2=s^2H(x_1,x_2), H ( s x 1 , s x 2 ) = 10 ( s x 1 ) ( s x 2 ) = s 2 ⋅ 10 x 1 x 2 = s 2 H ( x 1 , x 2 ) , so h ( f ( x 1 , x 2 ) ) h(f(x_1,x_2)) h ( f ( x 1 , x 2 )) homogeneous of degree 2 in ( x 1 , x 2 ) (x_1,x_2) ( x 1 , x 2 )
j ( y ) = y 2 j(y)=y^2 j ( y ) = y 2 k ( y ) = e y k(y)=e^y k ( y ) = e y
(a) j ( y ) = y 2 j(y)=y^2 j ( y ) = y 2
Homothetic? Yes on the given domain, because y = x 1 x 2 > 0 y=x_1x_2>0 y = x 1 x 2 > 0 y 2 y^2 y 2 y > 0 y>0 y > 0
Homogeneous in ( x 1 , x 2 ) (x_1,x_2) ( x 1 , x 2 ) Yes.J ( x 1 , x 2 ) = j ( f ( x 1 , x 2 ) ) = ( x 1 x 2 ) 2 J(x_1,x_2)=j(f(x_1,x_2))=(x_1x_2)^2 J ( x 1 , x 2 ) = j ( f ( x 1 , x 2 )) = ( x 1 x 2 ) 2
J ( s x 1 , s x 2 ) = ( ( s x 1 ) ( s x 2 ) ) 2 = ( s 2 x 1 x 2 ) 2 = s 4 ( x 1 x 2 ) 2 = s 4 J ( x 1 , x 2 ) , J(sx_1,sx_2)=\big((sx_1)(sx_2)\big)^2=(s^2x_1x_2)^2=s^4(x_1x_2)^2=s^4J(x_1,x_2), J ( s x 1 , s x 2 ) = ( ( s x 1 ) ( s x 2 ) ) 2 = ( s 2 x 1 x 2 ) 2 = s 4 ( x 1 x 2 ) 2 = s 4 J ( x 1 , x 2 ) , so it is homogeneous of degree 4 .
(b) k ( y ) = e y k(y)=e^y k ( y ) = e y
Homothetic? Yes, because e y e^y e y y y y
Homogeneous in ( x 1 , x 2 ) (x_1,x_2) ( x 1 , x 2 ) No.K ( x 1 , x 2 ) = k ( f ( x 1 , x 2 ) ) = e x 1 x 2 K(x_1,x_2)=k(f(x_1,x_2))=e^{x_1x_2} K ( x 1 , x 2 ) = k ( f ( x 1 , x 2 )) = e x 1 x 2
K ( s x 1 , s x 2 ) = e ( s x 1 ) ( s x 2 ) = e s 2 x 1 x 2 , K(sx_1,sx_2)=e^{(sx_1)(sx_2)}=e^{s^2x_1x_2}, K ( s x 1 , s x 2 ) = e ( s x 1 ) ( s x 2 ) = e s 2 x 1 x 2 , which cannot be written as s k K ( x 1 , x 2 ) s^k K(x_1,x_2) s k K ( x 1 , x 2 ) k k k
Summary
Function Homothetic? Homogeneous in ( x 1 , x 2 ) (x_1,x_2) ( x 1 , x 2 ) Degree f ( x 1 , x 2 ) = x 1 x 2 f(x_1,x_2)=x_1x_2 f ( x 1 , x 2 ) = x 1 x 2 — Yes 2 g ( y ) = ln y g(y)=\ln y g ( y ) = ln y Yes No — h ( y ) = 10 y h(y)=10y h ( y ) = 10 y Yes Yes 2 j ( y ) = y 2 j(y)=y^2 j ( y ) = y 2 Yes Yes 4 k ( y ) = e y k(y)=e^y k ( y ) = e y Yes No —
Every homogeneous function is homothetic , because a homogeneous function is already a special case of a monotonic transformation (the identity transformation).
However, the converse is not always true:
A homothetic function need not be homogeneous.
Monotonic transformations such as ln ( ⋅ ) \ln(\cdot) ln ( ⋅ ) e ( ⋅ ) e^{(\cdot)} e ( ⋅ ) destroy homogeneity , even though they preserve ordering.
Homogeneous ⟹ Homothetic , but not necessarily vice versa . \text{Homogeneous} \;\Longrightarrow\; \text{Homothetic},
\qquad
\text{but not necessarily vice versa}. Homogeneous ⟹ Homothetic , but not necessarily vice versa . This distinction is crucial in production theory and consumer theory:optimal input proportions , while homogeneous functions additionally impose scale properties .
Show that each of the following functions is homothetic by transforming it back to its underlying homogeneous form.
i. y = ln ( x ) + ln ( z ) y = \ln(x) + \ln(z) y = ln ( x ) + ln ( z )
ii. y = 0.3 ln ( L ) + 0.7 ln ( K ) y = 0.3 \ln(L) + 0.7 \ln(K) y = 0.3 ln ( L ) + 0.7 ln ( K )
iii. y = 2 ln ( x ) + ln ( y ) − ln ( w ) y = 2\ln(x) + \ln(y) - \ln(w) y = 2 ln ( x ) + ln ( y ) − ln ( w )
iv. y = e x z y = e^{xz} y = e x z
Answers ¶ i.y = ln ( x ) + ln ( z ) y = \ln(x) + \ln(z) y = ln ( x ) + ln ( z )
Exponentiating both sides givese y = x z e^{y} = xz e y = x z
The function x z xz x z ln ( ⋅ ) \ln(\cdot) ln ( ⋅ ) homothetic .
ii.y = 0.3 ln ( L ) + 0.7 ln ( K ) y = 0.3 \ln(L) + 0.7 \ln(K) y = 0.3 ln ( L ) + 0.7 ln ( K )
Exponentiating both sides givese y = L 0.3 K 0.7 e^{y} = L^{0.3} K^{0.7} e y = L 0.3 K 0.7
The Cobb–Douglas function L 0.3 K 0.7 L^{0.3} K^{0.7} L 0.3 K 0.7 homothetic .
iii.y = 2 ln ( x ) + ln ( y ) − ln ( w ) y = 2\ln(x) + \ln(y) - \ln(w) y = 2 ln ( x ) + ln ( y ) − ln ( w )
Exponentiating both sides givese y = x 2 y w e^{y} = \dfrac{x^{2} y}{w} e y = w x 2 y
The function x 2 y w \dfrac{x^{2} y}{w} w x 2 y homothetic .
iv.y = e x z y = e^{xz} y = e x z
Taking logs givesln ( y ) = x z \ln(y) = xz ln ( y ) = x z
The function x z xz x z y = e x z y = e^{xz} y = e x z homothetic .
Examples in Economics ¶ Consider a Cobb–Douglas production function
Q = A K 1 − α L α Q = A K^{1-\alpha} L^{\alpha} Q = A K 1 − α L α The tangent (slope) to its isoquant is
d K d L = ( α 1 − α ) K L \frac{dK}{dL}
=
\left( \frac{\alpha}{1-\alpha} \right)\frac{K}{L} d L d K = ( 1 − α α ) L K An interesting property of the Cobb–Douglas production function is that the ratio of its marginal products depends on the capital–labor ratio and not on the overall level of production.
Hence multiplying K K K L L L s s s
d K d L = ( α 1 − α ) K L = ( α 1 − α ) s K s L \frac{dK}{dL}
=
\left( \frac{\alpha}{1-\alpha} \right)\frac{K}{L}
=
\left( \frac{\alpha}{1-\alpha} \right)\frac{sK}{sL} d L d K = ( 1 − α α ) L K = ( 1 − α α ) s L sK More generally, the slope of the level curves of any homogeneous function is constant along any ray from the origin.
The slope of a level curve of a function f ( x 1 , x 2 ) f(x_1,x_2) f ( x 1 , x 2 )
d x 2 d x 1 = f 1 ( x 1 , x 2 ) f 2 ( x 1 , x 2 ) \frac{dx_2}{dx_1}
=
\frac{f_1(x_1,x_2)}{f_2(x_1,x_2)} d x 1 d x 2 = f 2 ( x 1 , x 2 ) f 1 ( x 1 , x 2 ) Now recall that for a homogeneous function of degree k k k
f i ( s x 1 , s x 2 ) = s k − 1 f i ( x 1 , x 2 ) f_i(sx_1,sx_2)
=
s^{k-1} f_i(x_1,x_2) f i ( s x 1 , s x 2 ) = s k − 1 f i ( x 1 , x 2 ) Thus along a ray from the origin,
d x 2 d x 1 = f 1 ( s x 1 , s x 2 ) f 2 ( s x 1 , s x 2 ) [ 0.5 e m ] = s k − 1 f 1 ( x 1 , x 2 ) s k − 1 f 2 ( x 1 , x 2 ) [ 0.5 e m ] = f 1 ( x 1 , x 2 ) f 2 ( x 1 , x 2 ) . \begin{aligned}
\frac{dx_2}{dx_1}
&=
\frac{f_1(sx_1,sx_2)}{f_2(sx_1,sx_2)} [0.5em]
&=
\frac{s^{k-1} f_1(x_1,x_2)}{s^{k-1} f_2(x_1,x_2)} [0.5em]
&=
\frac{f_1(x_1,x_2)}{f_2(x_1,x_2)} .
\end{aligned} d x 1 d x 2 = f 2 ( s x 1 , s x 2 ) f 1 ( s x 1 , s x 2 ) [ 0.5 e m ] = s k − 1 f 2 ( x 1 , x 2 ) s k − 1 f 1 ( x 1 , x 2 ) [ 0.5 e m ] = f 2 ( x 1 , x 2 ) f 1 ( x 1 , x 2 ) . Put differently, any proportional scaling s s s x 1 x_1 x 1 x 2 x_2 x 2
This property also extends to homothetic functions .
Consider the function
y = f ( x 1 , x 2 ) , y = f(x_1,x_2), y = f ( x 1 , x 2 ) , which we assume to be homogeneous of degree k k k
z = g ( y ) = g ( f ( x 1 , x 2 ) ) , z = g(y) = g(f(x_1,x_2)), z = g ( y ) = g ( f ( x 1 , x 2 )) , where g ′ ( y ) > 0 g'(y) > 0 g ′ ( y ) > 0 y y y g ′ ( y ) < 0 g'(y) < 0 g ′ ( y ) < 0 y y y
Using the chain rule together with the Implicit Function Theorem , we obtain
d x 2 d x 1 = g ′ ( y ) f 1 ( s x 1 , s x 2 ) g ′ ( y ) f 2 ( s x 1 , s x 2 ) = f 1 ( s x 1 , s x 2 ) f 2 ( s x 1 , s x 2 ) = s k − 1 f 1 ( x 1 , x 2 ) s k − 1 f 2 ( x 1 , x 2 ) = f 1 ( x 1 , x 2 ) f 2 ( x 1 , x 2 ) \begin{aligned}
\frac{dx_2}{dx_1}
&=
\frac{g'(y) f_1(sx_1,sx_2)}{g'(y) f_2(sx_1,sx_2)}
&=
\frac{f_1(sx_1,sx_2)}{f_2(sx_1,sx_2)}
&=
\frac{s^{k-1} f_1(x_1,x_2)}{s^{k-1} f_2(x_1,x_2)}
&=
\frac{f_1(x_1,x_2)}{f_2(x_1,x_2)}
\end{aligned} d x 1 d x 2 = g ′ ( y ) f 2 ( s x 1 , s x 2 ) g ′ ( y ) f 1 ( s x 1 , s x 2 ) = f 2 ( s x 1 , s x 2 ) f 1 ( s x 1 , s x 2 ) = s k − 1 f 2 ( x 1 , x 2 ) s k − 1 f 1 ( x 1 , x 2 ) = f 2 ( x 1 , x 2 ) f 1 ( x 1 , x 2 ) Thus, as with homogeneous functions, the scaling factor s s s any homothetic function —a class that includes but is not limited to homogeneous functions—is not altered by proportional scaling of all its arguments.
Lucas, R. E. (1990). Why Doesn’t Capital Flow from Rich to Poor Countries? American Economic Review , 80 (2), 92–96.