Implicit Differentiation

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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \left(0,0\right)} and radius 4, and is really the graph of two functions Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sin y-3x^{2}y=8} .

Note the $\sin y$ term requires the chain rule, the Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \left(1,0\right)} .

We first compute $y'$ by implicit differentiation.

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \left(1,0\right)} , we have $x=1$ and $y=0$. Plugging these into our equation for $y'$ gives

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \left(1,0\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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ye^{y}} to be Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ye^{y}y'+y'e^{y}} .

So

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}\neq0). \end{array}}

But we mustn't leave Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'} in our final answer. So, plug Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'=\dfrac{1}{e^{y}\left(y+1\right)}} back in to get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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.

~Page created by Jordan Tousignant