Implicit and inverse-function differentiation

Not every function comes prepackaged as $y = f(x)$. Many relations between $x$ and $y$ are implicit: $x^2 + y^2 = 1$, $x y + \sin y = x$, and so on. Implicit differentiation treats $y$ as a (possibly multi-valued) function of $x$, applies $\dfrac{d}{dx}$ to both sides of the equation, and solves for $\dfrac{dy}{dx}$.

A related technique uses the chain rule to find the derivative of an inverse function: if $y = f^{-1}(x)$, then $\dfrac{dy}{dx} = \dfrac{1}{f'(y)}$.

Implicit differentiation: the procedure

Given $F(x, y) = 0$:

1. Apply $\dfrac{d}{dx}$ to both sides, treating $y$ as an implicit function of $x$.
2. Use the chain rule for any function of $y$: $\dfrac{d}{dx}(y^n) = n y^{n - 1} \dfrac{dy}{dx}$, etc.
3. Collect $\dfrac{dy}{dx}$ terms on one side; solve algebraically.

Worked example: a circle

$x^2 + y^2 = 25$. Differentiate: $2x + 2y \dfrac{dy}{dx} = 0$, so $\dfrac{dy}{dx} = -\dfrac{x}{y}$.

At $(3, 4)$: $\dfrac{dy}{dx} = -3/4$.

Another example

$x y + \sin y = x$. Differentiate: $y + x \dfrac{dy}{dx} + \cos y \cdot \dfrac{dy}{dx} = 1$. Solve: $\dfrac{dy}{dx}(x + \cos y) = 1 - y$, so $\dfrac{dy}{dx} = \dfrac{1 - y}{x + \cos y}$.

Derivative of an inverse

If $y = f^{-1}(x)$, then $x = f(y)$. Differentiate both sides with respect to $x$: $1 = f'(y) \dfrac{dy}{dx}$, so

$\frac{dy}{dx} = \frac{1}{f'(y)} = \frac{1}{f'(f^{-1}(x))}.$

This is the inverse function theorem: the derivative of the inverse is the reciprocal of the derivative of the original (evaluated at the right point).

Application: derivatives of inverse trig

For $y = \sin^{-1} x$: $x = \sin y$, so $\dfrac{dy}{dx} = \dfrac{1}{\cos y} = \dfrac{1}{\sqrt{1 - \sin^2 y}} = \dfrac{1}{\sqrt{1 - x^2}}$ (positive square root, since $y \in [-\pi/2, \pi/2]$).

For $y = \tan^{-1} x$: $x = \tan y$, so $\dfrac{dy}{dx} = \dfrac{1}{\sec^2 y} = \dfrac{1}{1 + \tan^2 y} = \dfrac{1}{1 + x^2}$.

For $y = \cos^{-1} x$: $x = \cos y$ with $y \in [0, \pi]$, so $\sin y \ge 0$ and $\dfrac{dy}{dx} = -\dfrac{1}{\sin y} = -\dfrac{1}{\sqrt{1 - x^2}}$.

Worked examples

Example 1. Find $\dfrac{dy}{dx}$ from $x^3 + y^3 = 3 a x y$ (the folium of Descartes).

Differentiate: $3x^2 + 3 y^2 \dfrac{dy}{dx} = 3a(y + x \dfrac{dy}{dx})$. Rearrange: $\dfrac{dy}{dx}(y^2 - a x) = a y - x^2$, so $\dfrac{dy}{dx} = \dfrac{a y - x^2}{y^2 - a x}$.

Example 2. Find $\dfrac{dy}{dx}$ from $e^x + e^y = e^{x + y}$.

Differentiate: $e^x + e^y \dfrac{dy}{dx} = e^{x + y}(1 + \dfrac{dy}{dx})$. Rearrange: $\dfrac{dy}{dx}(e^y - e^{x + y}) = e^{x + y} - e^x$, so $\dfrac{dy}{dx} = \dfrac{e^{x + y} - e^x}{e^y - e^{x + y}} = \dfrac{e^x(e^y - 1)}{e^y(1 - e^x)} = -\dfrac{e^x(1 - e^y)}{e^y(1 - e^x)}$.

Simplify: $\dfrac{dy}{dx} = -\dfrac{e^x(e^y - 1)}{e^y(e^x - 1)}$ after cancellation.

Example 3. If $y = \cos^{-1}(\cos x)$, find $\dfrac{dy}{dx}$ on $(0, \pi)$.

On this interval $y = x$, so $\dfrac{dy}{dx} = 1$.

Example 4. If $y = \tan^{-1}\left(\dfrac{1 - \cos x}{\sin x}\right)$ for $x \in (0, \pi)$, find $\dfrac{dy}{dx}$.

Simplify the inside: $\dfrac{1 - \cos x}{\sin x} = \dfrac{2 \sin^2(x/2)}{2 \sin(x/2) \cos(x/2)} = \tan(x/2)$. So $y = \tan^{-1}(\tan(x/2)) = x/2$ on the principal branch. Hence $\dfrac{dy}{dx} = 1/2$.

Example 5. If $x = a \cos\theta, y = a \sin\theta$, find $\dfrac{dy}{dx}$.

Method 1: parametric (next subtopic). Method 2: $x^2 + y^2 = a^2$, so $\dfrac{dy}{dx} = -x/y = -\cos\theta/\sin\theta = -\cot\theta$.

Example 6. If $y = f^{-1}(x)$ and $f'(y) = 1/(1 + y^2)$, find $\dfrac{dy}{dx}$ in terms of $x$.

$\dfrac{dy}{dx} = 1/f'(y) = 1 + y^2$. This is general. (Here $f = \tan^{-1}$, so $y = \tan x$ and indeed $\dfrac{dy}{dx} = \sec^2 x = 1 + \tan^2 x = 1 + y^2$.)

Try it yourself

1. Find $\dfrac{dy}{dx}$ from $x^2 + y^2 = 16$ at $(2, 2\sqrt 3)$.
2. Find $\dfrac{dy}{dx}$ from $x y = 4$.
3. Find $\dfrac{dy}{dx}$ from $\sin(xy) = 1$ at any allowed point.
4. Find $\dfrac{dy}{dx}$ from $y = \log(\tan(x/2 + \pi/4))$.
5. Find $\dfrac{dy}{dx}$ from $y = \tan^{-1}\left(\dfrac{\sin x}{1 + \cos x}\right)$.
6. Derive the derivative of $\sec^{-1}$ using the inverse-function theorem.
7. Find $\dfrac{dy}{dx}$ from $x e^y + y e^x = 1$.
8. Find $\dfrac{dy}{dx}$ from $x^3 + y^3 - 3xy = 0$.
9. Find $\dfrac{dy}{dx}$ from $\tan^{-1}(y/x) = \log\sqrt{x^2 + y^2}$.
10. Find $\dfrac{dy}{dx}$ from $\cos^{-1}\left(\dfrac{1 - x^2}{1 + x^2}\right) = 2\tan^{-1} x$.
11. If $x \sqrt{1 + y} + y \sqrt{1 + x} = 0$, find $\dfrac{dy}{dx}$.
12. Find the slope of the tangent to $y^2 = 4x$ at the point $(1, 2)$.
13. If $\sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y)$, find $\dfrac{dy}{dx}$ (Hint: use $\sin^{-1}$ trick).
14. If $y = \sqrt{x + \sqrt{x + \sqrt{x + \ldots}}}$, find $\dfrac{dy}{dx}$. (Square both sides first.)

Pitfalls / Tricks

• Don't forget that derivatives of $y$ pick up a factor of $\dfrac{dy}{dx}$ , this is the chain rule applied implicitly.
• Solve algebraically for $\dfrac{dy}{dx}$ before plugging in numerical points.
• The inverse-function rule $1/f'(y)$ requires $f'(y) \neq 0$.
• For nested or trig-identity-laden expressions, look for a simplification before differentiating.

Next, logarithmic differentiation.