Implicit Differentiation

Background

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

Sometimes, functions described by equations in 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} and 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} are too hard to solve 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} , for example 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^{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 (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} explicitly. We do this by implicit differentiation. The process is to take the derivative of both sides of the given equation 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} , and then do some algebra steps to solve 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'} (or 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 \dfrac{dy}{dx}} if you prefer), keeping in mind that 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 a function 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 x} throughout the equation.

Warm-up exercises

Given that 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 a function 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 x} , find the derivative of the following functions 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} .

1. 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^{2}}

Solution: 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 2yy'}

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

2. 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 xy}

Solution: 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 xy'+y}

Reason: 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} and 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} are both functions 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 x} , and they are being multiplied together, so the product rule says it's 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\cdot y'+y\cdot1} .

3. 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 \cos y}

Solution: 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'\sin y}

Reason: The function 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 inside of the cosine function, so the chain rule gives 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 (-\sin y)y'=-y'\sin y} .

4. 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 \sqrt{x+y}}

Solution: 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 \frac{1+y'}{2\sqrt{x+y}}}

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

Exercise 1: 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 \sin y-3x^{2}y=8} .

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

We 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} \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 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 \tan y=\dfrac{y}{x}}   at the point   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)} .

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

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} \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 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)} , 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 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

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

Here's a picture of the curve and its 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

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

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.