Apply the product, quotient and chain rules, and differentiate exponential, logarithmic and trigonometric functions

What this dot point is asking

NESA wants you to differentiate any function built from the standard library (polynomials, $e^x$, $\ln x$, $\sin x$, $\cos x$, $\tan x$) using the power, chain, product and quotient rules. Almost every Maths Advanced calculus question begins with a differentiation step, so fluency here is non-negotiable.

The answer

The power rule

For any real $n$,

This extends to negative and fractional powers. For example, $\frac{d}{dx}(x^{-2}) = -2 x^{-3}$ and $\frac{d}{dx}(\sqrt{x}) = \frac{1}{2}x^{-1/2}$.

The chain rule

If $y = f(g(x))$, let $u = g(x)$ so $y = f(u)$. Then

In practice, "differentiate the outside, leave the inside alone, then multiply by the derivative of the inside".

The product rule

If $y = u(x) v(x)$, then

The quotient rule

If $y = \frac{u(x)}{v(x)}$, then

Standard derivatives

Memorise these. They appear in nearly every paper.

For composed trig, the chain rule gives $\frac{d}{dx}(\sin(f(x))) = f'(x) \cos(f(x))$, and similarly for $\cos$ and $\tan$.

Worked examples

Chain rule with trig

Differentiate $y = \sin(3x^2)$.

Let $u = 3x^2$, so $y = \sin u$. Then $\frac{dy}{du} = \cos u$ and $\frac{du}{dx} = 6x$.

$\frac{dy}{dx} = 6x \cos(3x^2).$

Product rule

Differentiate $y = x e^x$.

$u = x$, $v = e^x$, so $u' = 1$ and $v' = e^x$.

$\frac{dy}{dx} = 1 \cdot e^x + x \cdot e^x = e^x (1 + x).$

Quotient rule

Differentiate $y = \frac{\ln x}{x}$.

$u = \ln x$, $v = x$, $u' = \frac{1}{x}$, $v' = 1$.

$\frac{dy}{dx} = \frac{\frac{1}{x} \cdot x \; - \; \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}.$

Combined chain and product

Differentiate $y = x^2 \sin(2x)$.

Product rule with $u = x^2$, $v = \sin(2x)$. Then $u' = 2x$ and $v' = 2 \cos(2x)$ (chain rule on the inside).

$\frac{dy}{dx} = 2x \sin(2x) + 2 x^2 \cos(2x).$

Common traps

Forgetting the chain rule on composed functions. Writing $\frac{d}{dx}(\sin(2x)) = \cos(2x)$ loses the factor of $2$. The correct answer is $2 \cos(2x)$.

Mixing up the quotient rule sign. The numerator is $u' v$ minus $u v'$, in that order. Reversing it changes the sign.

Treating $e^{2x}$ like $2 e^{2x \log e}$. Just use the chain rule: $\frac{d}{dx}(e^{2x}) = 2 e^{2x}$.

Power rule on $a^x$. $\frac{d}{dx}(2^x)$ is not $x \cdot 2^{x - 1}$. For non-$e$ exponentials, write $2^x = e^{x \ln 2}$ first, giving $\frac{d}{dx}(2^x) = (\ln 2) \cdot 2^x$.

Not simplifying. Markers often reward a clean factored form. After the quotient rule, look for common factors.

In one sentence

Differentiation in Maths Advanced is the disciplined application of the power, chain, product and quotient rules to the standard derivatives of polynomial, exponential, logarithmic and trigonometric functions.