Chain Rule and Implicit Differentiation

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Objectives:

1.: Understand when to use the chain rule.
2.: Know how to use the chain rule to compute partial derivatives of the composition of functions.
3.: Understand how the formula for implicit differentiation follows from the chain rule.
4.: Be able to compute partials of implicit functions.

Recap Video

You can watch watch the following video which recaps the ideas of the section.

Chain Rule Recap

Implicit Differentiation Recap

Examples

Below is a video showing an example of using the chain rule.

If $f(x,y,z) = xe^{xyz}$ and $x=s$ , $y = \frac {1}{s+t}$ , and $z = t$ , evaluate $\displaystyle \frac {\partial f}{\partial s}$ and $\displaystyle \frac {\partial f}{\partial t}$ .

Below is a video showing an example of using implicit differentiation.

Find $\displaystyle \frac {\partial z}{\partial x}$ and $\displaystyle \frac {\partial z}{\partial y}$ when $x^2+y^2+z^2+\sin (4x)+\sin (4y)+\sin (4z) = 1$ .

Problems

Suppose $w= f(x,y,z) = xy+y^2z+2x$ , with $x = 2\cos (t)$ , $y = 2\sin (t)$ , $z = t$ . Evaluate $\frac {dw}{dt}$ at the point $(x,y,z) = (2,0,2\pi )$ .

The chain rule says

$\begin{eqnarray*} \frac{dw}{dt}&=& \frac{\partial w}{\partial x} \frac{\answer{dx}}{\answer{dt}}+ \frac{\partial w}{\partial y} \frac{\answer{dy}}{\answer{dt}}+ \frac{\partial w}{\partial z} \frac{\answer{dz}}{\answer{dt}} \\ &=& (\answer{y+2})(-2\sin(t)) + (x+2yz)\cdot \answer{2\cos(t)} + \answer{y^2} \cdot 1. \end{eqnarray*}$

The $t$ -value corresponding to the point $(x,y,z) = (2,0,2\pi )$ is $t=\answer {2\pi }$ . Plugging in $x = 2$ , $y = 0$ , $z=2\pi$ , and the above $t$ -value into $\frac {dw}{dt}$ gives

$\frac {dw}{dt}\Big |_{(2,0,2\pi )} = \answer {4}.$

Suppose $f(x,y,z) = 2x^2y^2+\sin (xy-1)+2xyz$ , and let $g(u,v) = (u^2-v^2,u+v,u^2+v^2)$ . Let $h = f(g(u,v))$ . Evaluate $\frac {\partial h}{\partial u}$ when $u = 1$ and $v = 0$ .

$\frac {\partial h}{\partial u}\Big |_{(u,v) = (1,0)} = \answer {25}$

The chain rule says

$\frac {\partial h}{\partial u}=\frac {\partial f}{\partial x}\frac {\partial x}{\partial u}+\frac {\partial f}{\partial y}\frac {\partial y}{\partial u}+\frac {\partial f}{\partial z}\frac {\partial z}{\partial u}.$

When $u=1$ and $v=0$ , we get $x = 1$ , $y = 1$ , and $z = 1$ . Plugging in all these values, we get

$\frac {\partial h}{\partial u}\Big |_{(u,v) = (1,0)} = 25.$

Suppose $x^2y^3+x^3z^3+y^2z^2 = 3$ . Evaluate $\frac {\partial z}{\partial x}$ at the point $(1,1,1)$ .

From the phrasing of the problem, we see that $x$ is a variable, $y$ is a variable, and $z$ is a variable. Let $F(x,y,z) = x^2y^3+x^3z^3+y^2z^2$ , so that the problem says

$F(x,y,z(x,y)) = 3.$ Using the chain rule to differentiate both sides of the equation with respect to $x$ gives

$\frac {\partial F}{\partial x} + \frac {\partial F}{\partial \answer {z}} \frac {\partial z}{\partial \answer {x}} = \answer {0}.$ We calculate:

$\frac {\partial F}{\partial x} = \answer {2xy^3+3x^2z^3}$

$\frac {\partial F}{\partial z} = \answer {3x^3z^2+2y^2z}$ Therefore,

$\frac {\partial z}{\partial x} = -\frac {F_x}{F_z} = -\frac {\answer {2xy^3+3x^2z^3}}{\answer {3x^3z^2+2y^2z}}.$ Plugging in $x=1$ , $y =1$ , and $z=1$ gives:

$\frac {\partial z}{\partial x}\Big |_{(1,1,1)} =\answer {-1}$

If $\sin (xyz) + \cos (x^2y^3z^2)=1$ , then $\displaystyle \frac {\partial x}{\partial y}$ at the point $(x,y,z) = (0,1,1)$ is equal to $\answer {0}$ .

Here, $x$ is the dependent variable, and $y$ and $z$ are the independent variables, so the equation reads $F(x(y,z),y,z) = 1$ .

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