Math Core

The chain rule is the single most-used rule in calculus: it powers implicit differentiation, inverse function derivatives, related rates, and later $u$ -substitution and the parametric slope $\dfrac{dy/dt}{dx/dt}$ . This lesson reviews all of it at BC speed.

The chain rule

Chain rule

If $y = f(g(x))$ , then

\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x), \qquad \text{or} \qquad \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}.

Differentiate the outer function (leaving the inside alone), then multiply by the derivative of the inside.

For deeper compositions, peel one layer at a time from the outside in, multiplying a factor for each layer. For $\sin^3(2x) = \big(\sin(2x)\big)^3$ , the layers are cube, sine, and $2x$ :

\frac{d}{dx}\sin^3(2x) = 3\sin^2(2x) \cdot \cos(2x) \cdot 2 = 6\sin^2(2x)\cos(2x).

Common chain-rule forms worth recognizing instantly:

\frac{d}{dx}e^{u} = e^{u}u', \qquad \frac{d}{dx}\ln u = \frac{u'}{u}, \qquad \frac{d}{dx}u^n = nu^{n-1}u', \qquad \frac{d}{dx}\sin u = u'\cos u.

Worked example: Chain rule from a table

Suppose $f(2) = -1$ , $f'(2) = 4$ , $g(1) = 2$ , $g'(1) = 3$ , and $f'(-1) = 6$ . Find the derivative at $x = 1$ of (a) $f(g(x))$ and (b) $\big[f(2x)\big]^3$ .

Solution. (a) $f'(g(1)) \cdot g'(1) = f'(2) \cdot 3 = 4 \cdot 3 = 12$ . Notice that $f'(-1)$ is a distractor: the outer derivative is evaluated at the inside value $g(1) = 2$ .

(b) Three layers: cube, $f$ , and $2x$ . The derivative is $3\big[f(2x)\big]^2 \cdot f'(2x) \cdot 2$ . At $x = 1$ : $3(-1)^2 \cdot 4 \cdot 2 = 24$ .

Common mistake

The outer derivative is evaluated at the inside, not at $x$ . Writing $f'(1)g'(1)$ for the derivative of $f(g(x))$ at $x = 1$ is the most common chain-rule error on table problems.

Implicit differentiation

When $x$ and $y$ are tangled together, differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ . Every $y$ term picks up a factor of $\dfrac{dy}{dx}$ by the chain rule, and products like $xy$ need the product rule. Then solve for $\dfrac{dy}{dx}$ . The result generally involves both $x$ and $y$ .

For the second derivative, differentiate $\dfrac{dy}{dx}$ again (quotient rule, with $y$ still a function of $x$ ), then substitute the expression for $\dfrac{dy}{dx}$ and simplify using the original equation when possible.

Worked example: An implicit second derivative

For $x^2 - y^2 = 1$ , find $\dfrac{d^2y}{dx^2}$ in terms of $y$ .

Solution. First derivative: $2x - 2y\,y' = 0$ , so $y' = \dfrac{x}{y}$ .

Second derivative by the quotient rule:

y'' = \frac{(1)(y) - x\,y'}{y^2} = \frac{y - x \cdot \frac{x}{y}}{y^2} = \frac{y^2 - x^2}{y^3}.

The original equation says $x^2 - y^2 = 1$ , so $y^2 - x^2 = -1$ and $y'' = -\dfrac{1}{y^3}$ .

Derivatives of inverse functions

If $g = f^{-1}$ and $f(a) = b$ , then $g(b) = a$ , and differentiating $f(g(x)) = x$ gives

Inverse function derivative

g'(b) = \frac{1}{f'(a)}, \qquad \text{where } f(a) = b.

Slopes of inverse functions at corresponding points are reciprocals.

The practical difficulty is finding $a$ : you are given the output $b$ and must find the input with $f(a) = b$ , usually by inspection (try small integers).

Worked example: Inverse slope without the inverse

Let $f(x) = x^3 + 2x + 1$ and $g = f^{-1}$ . Find $g'(4)$ .

Solution. Solve $f(a) = 4$ : try $a = 1$ , $1 + 2 + 1 = 4$ . ✓ Then $f'(x) = 3x^2 + 2$ , so $f'(1) = 5$ and $g'(4) = \dfrac{1}{5}$ .

Inverse trig derivatives

Applying the inverse rule to the trig functions gives

\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}, \qquad \frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^2}}, \qquad \frac{d}{dx}\arctan x = \frac{1}{1 + x^2}.

With the chain rule, $\dfrac{d}{dx}\arctan u = \dfrac{u'}{1 + u^2}$ and $\dfrac{d}{dx}\arcsin u = \dfrac{u'}{\sqrt{1 - u^2}}$ . These forms return in BC as antiderivatives and in power series for $\arctan x$ .

Logarithmic differentiation

For a variable base raised to a variable power, such as $y = x^x$ or $y = (\sin x)^{x}$ , neither the power rule nor the exponential rule applies. Take $\ln$ of both sides, differentiate implicitly, and multiply back by $y$ .

Worked example: A variable exponent

Find $\dfrac{dy}{dx}$ for $y = x^{x}$ , $x > 0$ , and evaluate it at $x = e$ .

Solution. $\ln y = x\ln x$ . Differentiate: $\dfrac{y'}{y} = \ln x + 1$ . So

y' = x^x(\ln x + 1).

At $x = e$ : $y' = e^e(1 + 1) = 2e^e$ .

Tip

Logarithmic differentiation also tames messy products and quotients: $\ln$ turns them into sums, which are much easier to differentiate.

Practice

Practice 1

Find $\dfrac{d}{dx}\left[\cos^4(3x)\right]$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 2

Find $\dfrac{d}{dx}\arctan\left(x^2\right)$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 3

Let $h(x) = f(g(x))$ , where $g(3) = -2$ , $g'(3) = 5$ , $f'(3) = 7$ and $f'(-2) = -2$ . Find $h'(3)$ .