Implicit Differentiation

Background

So far, you may only have differentiated functions written in the form $y=f\left(x\right)$. But some functions are better described by an equation involving $x$ and $y$. For example, $x^{2}+y^{2}=16$ describes the graph of a circle with center $\left(0,0\right)$ and radius 4, and is really the graph of two functions: $y=\pm {\sqrt {16-x^{2}}}$.

Sometimes, functions described by equations in $x$ and $y$ are too hard to solve for $y$, for example $x^{3}+y^{3}=6xy$. This equation really describes 3 different functions of x, whose graph is the curve:

We want to find derivatives of these functions without having to solve for $y$ explicitly. We can do this by implicit differentiation, in which we take the derivative of both sides of our equation with respect to $x$, and do some algebra steps to solve for $y'$ (or ${\dfrac {dy}{dx}}$ if you prefer), keeping in mind that $y$ is a function of $x$ throughout the equation.

Example 1

Find $y'$ if $\sin y-3x^{2}y=8$.

[Think $y=f\left(x\right)$ and momentarily view the equation as $\sin f\left(x\right)-3x^{2}f\left(x\right)=8$ to realize that the $\sin f\left(x\right)$ term requires the chain rule and the $3x^{2}f\left(x\right)$ term needs the product rule when differentiating, while the derivative of 8 is just 0.]

Then we get

${\begin{array}{rcl}\sin y-3x^{2}y&=&8\\\left(\cos y\right)\cdot y'-\left(3x^{2}y'+6xy\right)&=&0\quad {\text{(derivative of both sides with respect to x)}}\\\left(\cos y\right)\cdot y'-3x^{2}y'&=&6xy\\\left(\cos y-3x^{2}\right)y'&=&6xy\\y'&=&{\dfrac {6xy}{\cos y-3x^{2}}}\end{array}}$

Example 2

Find the equation of the tangent line to $\tan y={\dfrac {y}{x}}$ at the point $\left({\frac {\pi }{4}},{\frac {\pi }{4}}\right)$.

We first compute $y'$ by implicit differentiation. Note the derivative of the right side requires the quotient rule.

${\begin{array}{rcl}\tan y&=&{\dfrac {y}{x}}\\\left(\sec ^{2}y\right)\cdot y'&=&{\dfrac {xy'-y}{x^{2}}}\\x^{2}y'\cdot \sec ^{2}y&=&xy'-y\\x^{2}y'\cdot \sec ^{2}y-xy'&=&-y\\y'\left(x^{2}\sec ^{2}y-x\right)&=&-y\\y'&=&{\dfrac {-y}{x^{2}\sec ^{2}y-x}}\end{array}}$

At the point $\left({\frac {\pi }{4}},{\frac {\pi }{4}}\right)$ we have $x={\frac {\pi }{4}}$ and $y={\frac {\pi }{4}}$. Plugging these into our equation for $y'$ gives

${\begin{array}{rcl}y'&=&{\dfrac {-{\frac {\pi }{4}}}{\left({\frac {\pi }{4}}\right)^{2}\sec ^{2}\left({\frac {\pi }{4}}\right)-{\frac {\pi }{4}}}}\\&=&{\dfrac {-{\frac {\pi }{4}}}{{\frac {\pi ^{2}}{16}}\cdot 2-{\frac {\pi }{4}}}}\\&=&-{\dfrac {\pi }{4}}\cdot {\dfrac {8}{\pi ^{2}-2\pi }}\\&=&{\dfrac {2}{2-\pi }}\end{array}}$

This means the slope of the tangent line at $\left({\frac {\pi }{4}},{\frac {\pi }{4}}\right)$ is $m={\dfrac {2}{2-\pi }}$, and a point on this line is $\left({\frac {\pi }{4}},{\frac {\pi }{4}}\right)$. Using the point-slope form of a line, we have

${\begin{array}{rcl}y-{\frac {\pi }{4}}&=&{\frac {2}{2-\pi }}\left(x-{\frac {\pi }{4}}\right)\\y&=&{\frac {2}{2-\pi }}x-{\frac {\pi }{2\left(2-\pi \right)}}+{\frac {\pi }{4}}\\y&=&{\frac {2}{2-\pi }}x-{\frac {\pi ^{2}}{4\left(2-\pi \right)}}\end{array}}$

Example 3

Find $y''$ if $ye^{y}=x$.

Use implicit differentiation to find $y'$ first:

${\begin{array}{rcl}y\cdot e^{y}&=&x\\y\cdot e^{y}y'+y'\cdot e^{y}&=&1\\y'\left(ye^{y}+e^{y}\right)&=&1\\y'&=&{\dfrac {1}{ye^{y}+e^{y}}}\\&{\textrm {or}}&\left(ye^{y}+e^{y}\right)^{-1}\end{array}}$

Now $y''$ is just the derivative of $\left(ye^{y}+e^{y}\right)^{-1}$ with respect to $x$ (remember $y=f\left(x\right)$). This will require the chain rule. Note that we already found the derivative of $ye^{y}$ to be $ye^{y}y'+y'e^{y}$. So

${\begin{array}{rcl}y''&=&-1\left(ye^{y}+e^{y}\right)^{-2}\left(ye^{y}y'+y'e^{y}+e^{y}y'\right)\\&=&{\dfrac {-1}{\left(ye^{y}+e^{y}\right)^{2}}}\cdot \left(ye^{y}y'+2y'e^{y}\right)\\&=&-{\dfrac {y'e^{y}\left(y+2\right)}{\left(e^{y}\right)^{2}\left(y+1\right)^{2}}}\\&=&-{\dfrac {y'\left(y+2\right)}{e^{y}\left(y+1\right)^{2}}}\end{array}}$

But we mustn't leave $y'$ in our final answer. So, plug $y'={\dfrac {1}{e^{y}\left(y+1\right)}}$ back in to get

${\begin{array}{rcl}y''&=&-{\dfrac {{\frac {1}{e^{y}\left(y+1\right)}}\left(y+2\right)}{e^{y}\left(y+1\right)^{2}}}\\&=&-{\dfrac {y+2}{\left(e^{y}\right)^{2}\left(y+1\right)^{3}}}\end{array}}$

as our final answer.