Lesson notes

This lesson is a tour of the chain rule, also called differentiating a function of a function. For a composite $f(g(x))$ the rule is $\tfrac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$: differentiate the outer function, keep the inner one unchanged, then multiply by the derivative of the inner function. Below we work through algebraic, logarithmic, exponential, and trigonometric examples in the order the teacher presents them.

A linear expression raised to a power

Differentiate $(ax+b)^4$. Treat $ax+b$ as the inner function. Differentiating the power gives $4(ax+b)^3$, and the inner derivative is $\tfrac{d}{dx}(ax+b) = a$.

$\frac{d}{dx}(ax+b)^4 = 4(ax+b)^3 \cdot a = 4a\,(ax+b)^3$

Modulus of a quadratic

Differentiate $\lvert x^2 - 5\rvert$. Using $\tfrac{d}{dx}\lvert x\rvert = \tfrac{x}{\lvert x\rvert}$ with inner function $x^2-5$:

$\frac{d}{dx}\lvert x^2-5\rvert = \frac{x^2-5}{\lvert x^2-5\rvert}\cdot \frac{d}{dx}(x^2-5) = \frac{2x\,(x^2-5)}{\lvert x^2-5\rvert}$

Square root of a quadratic

Differentiate $\sqrt{4-x^2}$. The outer derivative of $\sqrt{u}$ is $\tfrac{1}{2\sqrt{u}}$, and the inner derivative is $\tfrac{d}{dx}(4-x^2) = -2x$.

$\frac{d}{dx}\sqrt{4-x^2} = \frac{1}{2\sqrt{4-x^2}}\cdot(-2x) = \frac{-x}{\sqrt{4-x^2}}$

A quadratic squared

Differentiate $(3x^2+6x+8)^2$. The outer derivative gives $2(3x^2+6x+8)$, and the inner derivative is $6x+6 = 6(x+1)$.

$\frac{d}{dx}(3x^2+6x+8)^2 = 2(3x^2+6x+8)\cdot 6(x+1) = 12(x+1)(3x^2+6x+8)$

Log of a log

Differentiate $\log(\log x)$. With $\tfrac{d}{dx}\log x = \tfrac{1}{x}$ and inner function $\log x$:

$\frac{d}{dx}\log(\log x) = \frac{1}{\log x}\cdot \frac{1}{x} = \frac{1}{x\log x}$

Exponential times a logarithm (product rule)

Differentiate $e^{2x}\log x$. This needs the product rule $\tfrac{d}{dx}(uv) = u\,\tfrac{dv}{dx} + v\,\tfrac{du}{dx}$, and the chain rule for $e^{2x}$ whose derivative is $2e^{2x}$.

$\frac{d}{dx}\big(e^{2x}\log x\big) = e^{2x}\cdot\frac{1}{x} + \log x\cdot 2e^{2x} = e^{2x}\left(\frac{1}{x} + 2\log x\right)$

Logarithm of a multiple

Differentiate $\log(3x)$. With inner function $3x$:

$\frac{d}{dx}\log(3x) = \frac{1}{3x}\cdot 3 = \frac{1}{x}$

A quadratic to the power three over two

Differentiate $(ax^2+bx+c)^{3/2}$. The outer derivative gives $\tfrac{3}{2}(ax^2+bx+c)^{1/2}$, and the inner derivative is $2ax+b$.

$\frac{d}{dx}(ax^2+bx+c)^{3/2} = \frac{3}{2}\,(ax^2+bx+c)^{1/2}\,(2ax+b)$

Trigonometric functions of a function

Sine of a multiple

Using $\tfrac{d}{dx}(\sin x)^n = n(\sin x)^{n-1}\cos x$, take $\sin 3x$:

$\frac{d}{dx}\sin 3x = \cos 3x \cdot 3 = 3\cos 3x$

Sine of cosine of a square

Differentiate $\sin(\cos x^2)$. Peel one layer at a time: the derivative of $\sin$, then of $\cos$, then of $x^2$.

$\frac{d}{dx}\sin(\cos x^2) = \cos(\cos x^2)\cdot(-\sin x^2)\cdot 2x = -2x\,\sin x^2\,\cos(\cos x^2)$

Two times root cot of a square

Differentiate $2\sqrt{\cot x^2}$. The constant $2$ stays outside; the square root gives $\tfrac{1}{2\sqrt{\cot x^2}}$, the derivative of $\cot$ is $-\csc^2$, and the inner derivative of $x^2$ is $2x$.

$\frac{d}{dx}\,2\sqrt{\cot x^2} = \frac{1}{\sqrt{\cot x^2}}\cdot(-\csc^2 x^2)\cdot 2x = \frac{-2x\,\csc^2 x^2}{\sqrt{\cot x^2}}$

Writing $\csc^2 x^2 = \tfrac{1}{\sin^2 x^2}$ and $\sqrt{\cot x^2} = \tfrac{\sqrt{\cos x^2}}{\sqrt{\sin x^2}}$, this simplifies to

$\frac{-2x}{\sin^2 x^2}\cdot\frac{\sqrt{\sin x^2}}{\sqrt{\cos x^2}} = \frac{-2x}{\sqrt{\sin x^2}\,\sqrt{\sin^2 x^2\,\cos x^2}}$

Using $2\sin\theta\cos\theta = \sin 2\theta$ and multiplying top and bottom by $\sqrt{2}$ gives the tidy form

$\frac{-2\sqrt{2}\,x}{\sin x^2\,\sqrt{\sin 2x^2}}$

Key takeaways

• The chain rule differentiates a composite as outer derivative times inner derivative: $\tfrac{d}{dx}f(g(x)) = f'(g(x))\,g'(x)$.
• It applies to powers, roots, modulus, logarithms, exponentials, and trigonometric functions alike, and you peel one layer at a time for deeply nested expressions.
• The chain rule combines naturally with the product rule, and trigonometric identities such as $2\sin\theta\cos\theta = \sin 2\theta$ help simplify the final answer.