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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y} are too hard to solve for Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y} is a function of $x$ throughout the equation.

Warm-up exercises

Given that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 $2f(x)f'(x)$ by the chain rule, but write it as $2yy'$.

2. $xy$

Solution: $xy'+y$

Reason: $x$ and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y} are both functions of $x$, and they 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y} is inside of the cosine function, so the chain rule gives $(-\sin y)y'=-y'\sin y$.

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 ${\frac {1}{2}}\left(x+y\right)^{-{\frac {1}{2}}}\left(1+y'\right)$, 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 $\tan y={\dfrac {y}{x}}$  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.

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

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 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'} 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   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 \left(\frac{\pi}{4},\frac{\pi}{4}\right)} . Using the point-slope form of a line, we have

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-\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}}

Here's a picture of the curve and its tangent line:

Exercise 3: Compute y"

Find 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''} if 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}=x} .

Use implicit differentiation to find 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'} first:

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} 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 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''} is just 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 \left(ye^{y}+e^{y}\right)^{-1}} with respect to 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 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}} \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.