# 4.5 The chain rule  (Page 5/9)

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## Implicit differentiation of a function of two or more variables

Suppose the function $z=f\left(x,y\right)$ defines $y$ implicitly as a function $y=g\left(x\right)$ of $x$ via the equation $f\left(x,y\right)=0.$ Then

$\frac{dy}{dx}=-\frac{\partial f\text{/}\partial x}{\partial f\text{/}\partial y}$

provided ${f}_{y}\left(x,y\right)\ne 0.$

If the equation $f\left(x,y,z\right)=0$ defines $z$ implicitly as a differentiable function of $x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y,$ then

$\frac{dz}{dx}=-\frac{\partial f\text{/}\partial x}{\partial f\text{/}\partial z}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{dz}{dy}=-\frac{\partial f\text{/}\partial y}{\partial f\text{/}\partial z}$

as long as ${f}_{z}\left(x,y,z\right)\ne 0.$

[link] is a direct consequence of [link] . In particular, if we assume that $y$ is defined implicitly as a function of $x$ via the equation $f\left(x,y\right)=0,$ we can apply the chain rule to find $dy\text{/}dx\text{:}$

$\begin{array}{ccc}\hfill \frac{d}{dx}f\left(x,y\right)& =\hfill & \frac{d}{dx}\left(0\right)\hfill \\ \hfill \frac{\partial f}{\partial x}·\frac{dx}{dx}+\frac{\partial f}{\partial y}·\frac{dy}{dx}& =\hfill & 0\hfill \\ \hfill \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}·\frac{dy}{dx}& =\hfill & 0.\hfill \end{array}$

Solving this equation for $dy\text{/}dx$ gives [link] . [link] can be derived in a similar fashion.

Let’s now return to the problem that we started before the previous theorem. Using [link] and the function $f\left(x,y\right)={x}^{2}+3{y}^{2}+4y-4,$ we obtain

$\begin{array}{c}\frac{\partial f}{\partial x}=2x\hfill \\ \frac{\partial f}{\partial y}=6y+4.\hfill \end{array}$

$\frac{dy}{dx}=-\frac{\partial f\text{/}\partial x}{\partial f\text{/}\partial y}=-\frac{2x}{6y+4}=-\frac{x}{3y+2},$

which is the same result obtained by the earlier use of implicit differentiation.

## Implicit differentiation by partial derivatives

1. Calculate $dy\text{/}dx$ if $y$ is defined implicitly as a function of $x$ via the equation $3{x}^{2}-2xy+{y}^{2}+4x-6y-11=0.$ What is the equation of the tangent line to the graph of this curve at point $\left(2,1\right)?$
2. Calculate $\partial z\text{/}\partial x$ and $\partial z\text{/}\partial y,$ given ${x}^{2}{e}^{y}-yz{e}^{x}=0.$
1. Set $f\left(x,y\right)=3{x}^{2}-2xy+{y}^{2}+4x-6y-11=0,$ then calculate ${f}_{x}$ and ${f}_{y}\text{:}$ $\begin{array}{c}{f}_{x}=6x-2y+4\hfill \\ {f}_{y}=-2x+2y-6.\hfill \end{array}$
The derivative is given by
$\frac{dy}{dx}=-\frac{\partial f\text{/}\partial x}{\partial f\text{/}\partial y}=-\frac{6x-2y+4}{-2x+2y-6}=\frac{3x-y+2}{x-y+3}.$

The slope of the tangent line at point $\left(2,1\right)$ is given by
${\frac{dy}{dx}|}_{\left(x,y\right)=\left(2,1\right)}=\frac{3\left(2\right)-1+2}{2-1+3}=\frac{7}{4}.$

To find the equation of the tangent line, we use the point-slope form ( [link] ):
$\begin{array}{ccc}\hfill y-{y}_{0}& =\hfill & m\left(x-{x}_{0}\right)\hfill \\ \hfill y-1& =\hfill & \frac{7}{4}\left(x-2\right)\hfill \\ \hfill y& =\hfill & \frac{7}{4}x-\frac{7}{2}+1\hfill \\ \hfill y& =\hfill & \frac{7}{4}x-\frac{5}{2}.\hfill \end{array}$

2. We have $f\left(x,y,z\right)={x}^{2}{e}^{y}-yz{e}^{x}.$ Therefore,
$\begin{array}{c}\frac{\partial f}{\partial x}=2x{e}^{y}-yz{e}^{x}\hfill \\ \frac{\partial f}{\partial y}={x}^{2}{e}^{y}-z{e}^{x}\hfill \\ \frac{\partial f}{\partial z}=\text{−}y{e}^{x}.\hfill \end{array}$

$\begin{array}{ccccccc}\begin{array}{cc}\hfill \frac{\partial z}{\partial x}& =-\frac{\partial f\text{/}\partial x}{\partial f\text{/}\partial y}\hfill \\ & =-\frac{2x{e}^{y}-yz{e}^{x}}{\text{−}y{e}^{x}}\hfill \\ & =\frac{2x{e}^{y}-yz{e}^{x}}{y{e}^{x}}\hfill \end{array}\hfill & & & \text{and}\hfill & & & \begin{array}{cc}\hfill \frac{\partial z}{\partial y}& =-\frac{\partial f\text{/}\partial y}{\partial f\text{/}\partial z}\hfill \\ & =-\frac{{x}^{2}{e}^{y}-z{e}^{x}}{\text{−}y{e}^{x}}\hfill \\ & =\frac{{x}^{2}{e}^{y}-z{e}^{x}}{y{e}^{x}}.\hfill \end{array}\hfill \end{array}$

Find $dy\text{/}dx$ if $y$ is defined implicitly as a function of $x$ by the equation ${x}^{2}+xy-{y}^{2}+7x-3y-26=0.$ What is the equation of the tangent line to the graph of this curve at point $\left(3,-2\right)?$

$\frac{dy}{dx}={\frac{2x+y+7}{2y-x+3}|}_{\left(3,-2\right)}=\frac{2\left(3\right)+\left(-2\right)+7}{2\left(-2\right)-\left(3\right)+3}=-\frac{11}{4}$
Equation of the tangent line: $y=-\frac{11}{4}x+\frac{25}{4}$

## Key concepts

• The chain rule for functions of more than one variable involves the partial derivatives with respect to all the independent variables.
• Tree diagrams are useful for deriving formulas for the chain rule for functions of more than one variable, where each independent variable also depends on other variables.

## Key equations

• Chain rule, one independent variable
$\frac{dz}{dt}=\frac{\partial z}{\partial x}·\frac{dx}{dt}+\frac{\partial z}{\partial y}·\frac{dy}{dt}$
• Chain rule, two independent variables
$\frac{dz}{du}=\frac{\partial z}{\partial x}·\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}·\frac{\partial y}{\partial u}$
$\frac{dz}{dv}=\frac{\partial z}{\partial x}·\frac{\partial x}{\partial v}+\frac{\partial z}{\partial y}·\frac{\partial y}{\partial v}$
• Generalized chain rule
$\frac{\partial w}{\partial {t}_{j}}=\frac{\partial w}{\partial {x}_{1}}\phantom{\rule{0.2em}{0ex}}\frac{\partial {x}_{1}}{\partial {t}_{j}}+\frac{\partial w}{\partial {x}_{2}}\phantom{\rule{0.2em}{0ex}}\frac{\partial {x}_{1}}{\partial {t}_{j}}+\text{⋯}+\frac{\partial w}{\partial {x}_{m}}\phantom{\rule{0.2em}{0ex}}\frac{\partial {x}_{m}}{\partial {t}_{j}}$

For the following exercises, use the information provided to solve the problem.

Let $w\left(x,y,z\right)=xy\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}z,$ where $x=t,y={t}^{2},$ and $z=\text{arcsin}\phantom{\rule{0.2em}{0ex}}t.$ Find $\frac{dw}{dt}.$

$\frac{dw}{dt}=y\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}z+x\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}z\left(2t\right)-\frac{xy\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}z}{\sqrt{1-{t}^{2}}}$

Let $w\left(t,v\right)={e}^{tv}$ where $t=r+s$ and $v=rs.$ Find $\frac{\partial w}{\partial r}$ and $\frac{\partial w}{\partial s}.$

If $w=5{x}^{2}+2{y}^{2},x=-3s+t,$ and $y=s-4t,$ find $\frac{\partial w}{\partial s}$ and $\frac{\partial w}{\partial t}.$

$\frac{\partial w}{\partial s}=-30x+4y,$ $\frac{\partial w}{\partial t}=10x-16y$

If $w=x{y}^{2},x=5\phantom{\rule{0.2em}{0ex}}\text{cos}\left(2t\right),$ and $y=5\phantom{\rule{0.2em}{0ex}}\text{sin}\left(2t\right),$ find $\frac{\partial w}{\partial t}.$

If $f\left(x,y\right)=xy,x=r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta ,$ and $y=r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta ,$ find $\frac{\partial f}{\partial r}$ and express the answer in terms of $r$ and $\theta .$

$\frac{\partial f}{\partial r}=r\phantom{\rule{0.2em}{0ex}}\text{sin}\left(2\theta \right)$

Suppose $f\left(x,y\right)=x+y,u={e}^{x}\text{sin}\phantom{\rule{0.2em}{0ex}}y,x={t}^{2},$ and $y=\pi t,$ where $x=r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta$ and $y=r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta .$ Find $\frac{\partial f}{\partial \theta }.$

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