Apply the product, quotient and chain rules, including in combination, to differentiate functions built from polynomial, exponential, logarithmic and trigonometric components

What this dot point is asking

QCAA wants you to differentiate any function built from the standard library (polynomials, $e^x$, $\ln x$, $\sin x$, $\cos x$, $\tan x$) using combinations of the product, quotient and chain rules. Almost every Methods calculus question starts with a derivative step, so fluency here is non-negotiable for IA2, Paper 1 of the EA, and the calculus content of every PSMT.

The answer

The chain rule

If $y = f(g(x))$, set $u = g(x)$ 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 $y = \dfrac{u(x)}{v(x)}$ with $v(x) \neq 0$, then

The numerator is $u' v$ minus $u v'$, in that order. Reversing the order changes the sign.

Standard derivatives (the library)

The rules above act on the standard derivatives. Memorise these.

Order of operations

When two or more rules combine, choose the outer structure first and work inwards.

• IMATH_19 is a product; differentiate with the product rule, and the inside of $\sin x$ is just $x$ so no chain rule on the trig.
• IMATH_22 is a composition; differentiate with the chain rule, no product rule.
• IMATH_23 is a product whose factors each need the chain rule; apply the product rule and the chain rule lives inside each derivative.

Worked examples

Chain inside a product

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

Product rule with $u = x$, $v = \sin(2x)$, $u' = 1$, $v' = 2 \cos(2x)$ (chain on the inside).

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

Chain inside a quotient

Differentiate $y = \dfrac{e^{2x}}{x + 1}$.

Quotient rule with $u = e^{2x}$, $v = x + 1$, $u' = 2 e^{2x}$, $v' = 1$.

$\dfrac{dy}{dx} = \dfrac{2 e^{2x} (x + 1) - e^{2x} \cdot 1}{(x + 1)^2} = \dfrac{e^{2x} (2x + 1)}{(x + 1)^2}.$

Double chain

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

Outer power, then inner sine, then innermost $3x$.

$\dfrac{dy}{dx} = 2 \sin(3x) \cdot 3 \cos(3x) = 6 \sin(3x) \cos(3x) = 3 \sin(6x).$

The last step uses the double-angle identity $2 \sin\theta \cos\theta = \sin(2\theta)$.

Logarithm with the product rule

Differentiate $y = x \ln x$.

Product rule with $u = x$, $v = \ln x$, $u' = 1$, $v' = \tfrac{1}{x}$.

$\dfrac{dy}{dx} = \ln x + x \cdot \dfrac{1}{x} = \ln x + 1.$

Quotient with a chain inside

Differentiate $y = \dfrac{\cos(3x)}{x^2}$.

$u = \cos(3x)$, $v = x^2$, $u' = -3 \sin(3x)$, $v' = 2x$.

$\dfrac{dy}{dx} = \dfrac{-3 \sin(3x) \cdot x^2 - \cos(3x) \cdot 2x}{x^4} = \dfrac{-3 x \sin(3x) - 2 \cos(3x)}{x^3}.$

Common traps

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

Reversing the quotient rule sign. The numerator is $u' v$ minus $u v'$. Writing it the other way around flips the sign of the entire derivative.

Treating $e^{2x}$ like a power. $\frac{d}{dx}(e^{2x}) = 2 e^{2x}$, by the chain rule. It is not $2x \, e^{2x - 1}$.

Applying the power rule to $a^x$. For non-$e$ exponentials, $\frac{d}{dx}(a^x) = (\ln a) \, a^x$. The power rule does not apply because the base is constant and the exponent is variable.

Not simplifying. QCAA frequently allocates a mark for a clean final form. After a product or quotient rule, look for common factors and factor them out.

Choosing the wrong outer structure. $y = e^{x^2 + x}$ is a composition, not a product. Apply the chain rule, not the product rule.

In one sentence

The product rule gives $(uv)' = u'v + uv'$, the quotient rule gives $(u/v)' = (u'v - uv')/v^2$, and the chain rule gives $(f(g(x)))' = f'(g(x)) g'(x)$, and they combine in a clear inside-to-outside order whenever a function is built from polynomial, exponential, logarithmic and trigonometric pieces.