Chain Rule

The rule

The chain rule differentiates a composite function, one function applied to the output of another. If $y = f(g(x))$ , then

\dfrac{dy}{dx} = f'(g(x)) \cdot g'(x)

In words: differentiate the outer function (leaving the inner function alone), then multiply by the derivative of the inner function.

1. Derivative of sin(ln(x))

Find the derivative of $f(x) = \sin(\ln(x))$ .

Solution: The outer function is $\sin(u)$ , the inner function is $u = \ln(x)$ . Since $\dfrac{d}{du}\sin(u) = \cos(u)$ and $\dfrac{d}{dx}\ln(x) = \dfrac{1}{x}$ ,

f'(x) = \cos(\ln(x)) \cdot \dfrac{1}{x} = \dfrac{\cos(\ln(x))}{x}

2. Derivative of sec(tan(√x))

Find the derivative of $\sec(\tan(\sqrt{x}))$ with respect to $x$ . This nests three functions, so the chain rule is applied twice.

Solution: Let $u = \sqrt{x}$ , $v = \tan(u)$ , and differentiate $\sec(v)$ from the outside in:

\dfrac{du}{dx} = \dfrac{1}{2\sqrt{x}}, \qquad \dfrac{dv}{du} = \sec^2(u), \qquad \dfrac{d}{dv}\sec(v) = \sec(v)\tan(v)

Multiplying the three pieces together (the chain rule again):

\dfrac{1}{2\sqrt{x}} \cdot \sec(\tan(\sqrt{x})) \cdot \tan(\tan(\sqrt{x})) \cdot \sec^2(\sqrt{x})

Chain Rule

The rule

The chain rule differentiates a composite function, one function applied to the output of another. If $y = f(g(x))$ , then

\dfrac{dy}{dx} = f'(g(x)) \cdot g'(x)

In words: differentiate the outer function (leaving the inner function alone), then multiply by the derivative of the inner function.

1. Derivative of sin(ln(x))

Find the derivative of $f(x) = \sin(\ln(x))$ .

Solution: The outer function is $\sin(u)$ , the inner function is $u = \ln(x)$ . Since $\dfrac{d}{du}\sin(u) = \cos(u)$ and $\dfrac{d}{dx}\ln(x) = \dfrac{1}{x}$ ,

f'(x) = \cos(\ln(x)) \cdot \dfrac{1}{x} = \dfrac{\cos(\ln(x))}{x}

2. Derivative of sec(tan(√x))

Find the derivative of $\sec(\tan(\sqrt{x}))$ with respect to $x$ . This nests three functions, so the chain rule is applied twice.

Solution: Let $u = \sqrt{x}$ , $v = \tan(u)$ , and differentiate $\sec(v)$ from the outside in:

\dfrac{du}{dx} = \dfrac{1}{2\sqrt{x}}, \qquad \dfrac{dv}{du} = \sec^2(u), \qquad \dfrac{d}{dv}\sec(v) = \sec(v)\tan(v)

Multiplying the three pieces together (the chain rule again):

\dfrac{1}{2\sqrt{x}} \cdot \sec(\tan(\sqrt{x})) \cdot \tan(\tan(\sqrt{x})) \cdot \sec^2(\sqrt{x})

Chain Rule

The rule

1. Derivative of sin(ln(x))

2. Derivative of sec(tan(√x))

Continue learning

Chain Rule

The rule

1. Derivative of sin(ln(x))

2. Derivative of sec(tan(√x))

Continue learning