Implicit differentiation

So far, every function has had the form $y=f(x)$. The variable $y$ is alone on one side. We say $y$ is given explicitly.

Some curves are described by an equation in $x$ and $y$ instead. For example, the circle

$x^2+y^2=25$.

Here $y$ is given implicitly: the equation connects $x$ and $y$, but $y$ is not alone.

We could solve for $y$: $y=\pm\sqrt{25-x^2}$. But that gives two functions, and the square roots make the derivative messy. For other curves, such as $x^3+y^3=6xy$, solving for $y$ is very hard.

Implicit differentiation finds $\dfrac{dy}{dx}$ without solving for $y$.

The idea

We think of $y$ as a function of $x$, even though we do not know its formula. Then we differentiate both sides of the equation with respect to $x$.

The key step is differentiating a term with $y$ in it. Because $y$ is a function of $x$, we must use the chain rule. For example, $y^2$ is a square with $y$ inside. So

$\dfrac{d}{dx}(y^2)=2y\cdot\dfrac{dy}{dx}$.

Compare this with $\dfrac{d}{dx}(x^2)=2x$. The extra factor $\dfrac{dy}{dx}$ is the derivative of the inner function, $y$.

In general, every time we differentiate a term with $y$, we multiply by $\dfrac{dy}{dx}$:

$\dfrac{d}{dx}(y)=\dfrac{dy}{dx}$
$\dfrac{d}{dx}(y^n)=ny^{n-1}\dfrac{dy}{dx}$
$\dfrac{d}{dx}(\sin y)=\cos y\,\dfrac{dy}{dx}$

To keep the lines short, we often write $y'$ for $\dfrac{dy}{dx}$.

A common mistake: writing $\dfrac{d}{dx}(y^2)=2y$. This forgets the factor $y'$.

How to differentiate implicitly

Differentiate both sides of the equation with respect to $x$. Multiply by $y'$ each time you differentiate a term with $y$.
Move all the terms with $y'$ to one side, and all other terms to the other side.
Factor out $y'$.
Divide to solve for $y'$.

Example 1: For the circle $x^2+y^2=25$:

(a) Find $\dfrac{dy}{dx}$.

(b) Find the equation of the tangent line at $(3,4)$.

Solution:

(a) Differentiate both sides with respect to $x$:

$\dfrac{d}{dx}(x^2)+\dfrac{d}{dx}(y^2)=\dfrac{d}{dx}(25)$

$\begin{align*}2x+2yy'&=0\\2yy'&=-2x\\y'&=-\dfrac{x}{y}\end{align*}$

(b) The point $(3,4)$ is on the circle, because $3^2+4^2=25$.

The slope at $(3,4)$:

$y'=-\dfrac{3}{4}$

The tangent line:

$\begin{align*}y-4&=-\tfrac{3}{4}(x-3)\\y&=-\tfrac{3}{4}x+\tfrac{25}{4}\end{align*}$

The circle $x^2+y^2=25$ and its tangent line at $(3,4)$.

Check: Near $(3,4)$, the circle is the top half, $y=\sqrt{25-x^2}$. By the chain rule,

$y'=\dfrac{-x}{\sqrt{25-x^2}}$.

At $x=3$, this is $\dfrac{-3}{\sqrt{16}}=-\dfrac{3}{4}$. The answers match.

Notice that the slope $y'=-\dfrac{x}{y}$ uses both $x$ and $y$. This is normal in implicit differentiation. To find a slope, you need the whole point, not only $x$. For example, at $(3,-4)$ the slope is $-\dfrac{3}{-4}=\dfrac{3}{4}$.

Example 2: Find the slope of the tangent line to

$x^3+y^3=6xy$

at the point $(3,3)$.

Solution:

First, check the point: $27+27=54$ and $6(3)(3)=54$. So $(3,3)$ is on the curve.

Differentiate both sides. The right side, $6xy$, is a product, so use the product rule:

$\begin{align*}&\dfrac{d}{dx}(6xy)\\&=6(1\cdot y+x\cdot y')\\&=6y+6xy'\end{align*}$

So the equation becomes

$3x^2+3y^2y'=6y+6xy'$.

Move the $y'$ terms to the left, and the other terms to the right:

$3y^2y'-6xy'=6y-3x^2$

Factor out $y'$:

$y'(3y^2-6x)=6y-3x^2$

Divide both sides by $3y^2-6x$:

$y'=\dfrac{6y-3x^2}{3y^2-6x}$

Divide the numerator and the denominator by $3$:

$y'=\dfrac{2y-x^2}{y^2-2x}$

At $(3,3)$:

$y'=\dfrac{2(3)-9}{9-2(3)}=\dfrac{-3}{3}=-1$

The curve $x^3+y^3=6xy$ and its tangent line at $(3,3)$.

Example 3: Find $y'$ for

$x+\sin y=xy$.

Solution:

Differentiate each term:

$\dfrac{d}{dx}(x)=1$.
$\dfrac{d}{dx}(\sin y)=\cos y\cdot y'$, by the chain rule.
$\dfrac{d}{dx}(xy)=y+xy'$, by the product rule.

So the equation becomes

$1+\cos y\cdot y'=y+xy'$.

Solve for $y'$:

$\begin{align*}\cos y\cdot y'-xy'&=y-1\\y'(\cos y-x)&=y-1\\y'&=\dfrac{y-1}{\cos y-x}\end{align*}$

As always with trigonometric functions, $y$ is in radians.

Horizontal and vertical tangent lines

When $y'$ is a fraction, it tells us two things:

The tangent line is horizontal where the numerator is $0$ (and the denominator is not $0$). The slope is $0$ there.
The tangent line is vertical where the denominator is $0$ (and the numerator is not $0$). The slope is undefined there.

For example, the circle in Example 1 has vertical tangent lines at $(5,0)$ and $(-5,0)$. There, $y'=-\dfrac{x}{y}$ has the denominator $0$.

Example 4: Find the points on the curve

$x^2+xy+y^2=3$

where the tangent line is horizontal or vertical.

Solution:

Differentiate both sides. The middle term $xy$ needs the product rule:

$2x+(y+xy')+2yy'=0$

Solve for $y'$:

$\begin{align*}xy'+2yy'&=-2x-y\\y'(x+2y)&=-(2x+y)\\y'&=-\dfrac{2x+y}{x+2y}\end{align*}$

Horizontal: The numerator is $0$ when

$2x+y=0$, that is, $y=-2x$.

Put this into the equation of the curve:

$\begin{align*}x^2+x(-2x)+(-2x)^2&=3\\x^2-2x^2+4x^2&=3\\3x^2&=3\\x&=\pm 1\end{align*}$

Then $y=-2x$ gives the points $(1,-2)$ and $(-1,2)$.

Vertical: The denominator is $0$ when

$x+2y=0$, that is, $x=-2y$.

Put this into the equation of the curve:

$\begin{align*}(-2y)^2+(-2y)y+y^2&=3\\4y^2-2y^2+y^2&=3\\3y^2&=3\\y&=\pm 1\end{align*}$

Then $x=-2y$ gives the points $(-2,1)$ and $(2,-1)$.

At all four points, only one of the numerator and the denominator is $0$.

The curve $x^2+xy+y^2=3$. The tangent line is horizontal at two points and vertical at two points.

The power rule for fraction exponents

In basic differentiation rules, we said the power rule also works when the exponent is a fraction. Implicit differentiation shows why.

Let $y=x^{p/q}$, where $p$ and $q$ are integers, and $q$ is positive. Raise both sides to the power $q$:

$y^q=x^p$

Differentiate both sides implicitly:

$qy^{q-1}y'=px^{p-1}$

Divide both sides by $qy^{q-1}$:

$y'=\dfrac{p}{q}\cdot\dfrac{x^{p-1}}{y^{q-1}}$

Now replace $y$ by $x^{p/q}$. The denominator becomes

$y^{q-1}=x^{p(q-1)/q}=x^{p-p/q}$.

$\begin{align*}&y'\\&=\dfrac{p}{q}\cdot\dfrac{x^{p-1}}{x^{p-p/q}}\end{align*}$

Subtract the exponents:

$\begin{align*}&=\dfrac{p}{q}\,x^{(p-1)-(p-p/q)}\end{align*}$

Simplify the exponent:

$\begin{align*}&\dfrac{d}{dx}\left(x^{p/q}\right)=\dfrac{p}{q}\,x^{p/q-1}\end{align*}$

This is exactly the power rule with $n=\dfrac{p}{q}$.

Example 5: Find each derivative.

(a) $y=x^{3/4}$

(b) $y=\sqrt[3]{x^2+1}$

Solution:

(a) $y'=\dfrac{3}{4}x^{-1/4}$

(b) Rewrite the root as a power:

$y=(x^2+1)^{1/3}$

Inner function: $u=x^2+1$, so $u'=2x$.

$\begin{align*}&y'\\&=\tfrac{1}{3}(x^2+1)^{-2/3}\cdot 2x\\&=\dfrac{2x}{3(x^2+1)^{2/3}}\end{align*}$

Summary

Differentiate both sides with respect to $x$.
Each time you differentiate a term with $y$, multiply by $y'$ (the chain rule).
Collect the $y'$ terms, factor out $y'$, and solve.
The slope $y'$ usually depends on both $x$ and $y$, so you need the whole point.
Numerator $0$: horizontal tangent. Denominator $0$: vertical tangent.

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