Implicit Differentiation

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Background

So far, you may only have differentiated functions written in the form $y=f(x)$ . 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 $y=\pm {\sqrt {16-x^{2}}}$ , the upper and lower semicircles:

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 do this by implicit differentiation. The process is to take the derivative of both sides of the given equation with respect to $x$ , and then 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.

Warm-up exercises

Given that $y$ is a function of $x$ , find the derivative of the following functions with respect to $x$ .

1)   $y^{2}$ Solution:
$2yy'$ Reason:
Think $y=f(x)$ , and view it as $(f(x))^{2}$ to see that the derivative is $2f(x)\cdot f'(x)$ by the chain rule, but write it as $2yy'$ .

2)   $xy$ Solution:
$xy'+y$ Reason:
$x$ and $y$ are both functions of $x$ which are being multiplied together, so the product rule says it's $x\cdot y'+y\cdot 1$ .

3)   $\cos y$ Solution:
$-y'\sin y$ Reason:
The function $y$ is inside of the cosine function, so the chain rule gives $(-\sin y)\cdot y'=-y'\sin y$ .

4)   ${\sqrt {x+y}}$ Solution:
${\frac {1+y'}{2{\sqrt {x+y}}}}$ Reason:
Write it as $(x+y)^{\frac {1}{2}}$ , and use the chain rule to get   ${\frac {1}{2}}\left(x+y\right)^{-{\frac {1}{2}}}\cdot \left(1+y'\right)$ , then simplify.

Exercise 1: Compute y'

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

Note the $\sin y$ term requires the chain rule, the $3x^{2}y$ term needs the product rule, and the derivative of 8 is 0.

We get

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

Find the equation of the tangent line to $x^{2}+2xy-y^{2}+x=2$ at the point $\left(1,0\right)$ .

We first compute $y'$ by implicit differentiation.

${\begin{array}{rcl}x^{2}+2xy-y^{2}+x&=&2\\2x+2xy'+2y-2yy'+1&=&0\\x+xy'+y-yy'+{\frac {1}{2}}&=&0\\xy'-yy'&=&-x-y-{\frac {1}{2}}\\(x-y)y'&=&-(x+y+{\frac {1}{2}})\\y'&=&-{\dfrac {x+y+{\frac {1}{2}}}{x-y}}\end{array}}$ At the point $\left(1,0\right)$ , we have $x=1$ and $y=0$ . Plugging these into our equation for $y'$ gives

${\begin{array}{rcl}y'&=&-{\dfrac {1+0+{\frac {1}{2}}}{1-0}}=-{\frac {3}{2}}.\\\end{array}}$ This means the slope of the tangent line at $\left(0,1\right)$ is $m=-{\frac {3}{2}}$ , and a point on this line is $\left(1,0\right)$ . Using the point-slope form of a line, we get

${\begin{array}{rcl}y-0&=&-{\frac {3}{2}}\left(x-1\right)\\\\y&=&-{\frac {3}{2}}x+{\frac {3}{2}}.\\\end{array}}$ Here's a picture of the curve and tangent line:

Exercise 3: Compute y"

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

Use implicit differentiation to find $y'$ first:

${\begin{array}{rcl}ye^{y}&=&x\\ye^{y}y'+y'e^{y}&=&1\\y'\left(ye^{y}+e^{y}\right)&=&1\\y'&=&{\dfrac {1}{ye^{y}+e^{y}}}\\&=&\left(ye^{y}+e^{y}\right)^{-1}\end{array}}$ Now $y''$ is just the derivative of $\left(ye^{y}+e^{y}\right)^{-1}$ with respect to $x$ . This will require the chain rule. Notice we already found the derivative of $ye^{y}$ to be $ye^{y}y'+y'e^{y}$ .

So

${\begin{array}{rcl}y''&=&-1\left(ye^{y}+e^{y}\right)^{-2}\left(ye^{y}y'+y'e^{y}+e^{y}y'\right)\\\\&=&{\dfrac {-1}{\left(ye^{y}+e^{y}\right)^{2}}}\left(ye^{y}y'+2y'e^{y}\right)\\\\&=&-{\dfrac {y'e^{y}\left(y+2\right)}{\left(e^{y}\right)^{2}\left(y+1\right)^{2}}}\\\\&=&-{\dfrac {y'\left(y+2\right)}{e^{y}\left(y+1\right)^{2}}}\quad ({\text{since }}e^{y}\neq 0).\end{array}}$ 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

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