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 $\left(0,0\right)$ 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 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} which 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)\cdot 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}}\cdot\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 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=\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

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{\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}\cdot2-\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   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)} 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 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.