The product, quotient and chain rules of differentiation, and the derivatives of standard functions $x^n$ for $n \in Q$, $e^x$, $\ln(x)$, $\sin(x)$, $\cos(x)$ and $\tan(x)$

What this dot point is asking

VCAA wants fluent by-hand differentiation of any function built from the standard library (polynomials, $e^x$, $\ln x$, $\sin x$, $\cos x$, $\tan x$) using the four standard rules. Paper 1 almost always opens with a differentiation question and rewards clean, factored answers.

Standard derivatives

Memorise these. They appear in nearly every paper.

The power rule $\frac{d}{dx}(x^n) = n x^{n-1}$ extends to all rational $n$: e.g. $\frac{d}{dx}(\sqrt{x}) = \frac{1}{2} x^{-1/2}$, and $\frac{d}{dx}(x^{-2}) = -2 x^{-3}$.

The derivative of $\ln(x)$ is $\frac{1}{x}$ for $x > 0$. For the extended form, $\frac{d}{dx}(\ln|x|) = \frac{1}{x}$ for $x \neq 0$.

The sum rule

You differentiate term by term. Constants come out: $\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)$.

The product rule

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

In words: derivative of the first times the second, plus the first times derivative of the second.

Example. Differentiate $y = x^2 e^x$.

$u = x^2$, $v = e^x$. $u' = 2 x$, $v' = e^x$.

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

The quotient rule

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

Note the sign: $u' v$ minus $u v'$ in that order.

Example. Differentiate $y = \frac{x}{\ln(x)}$ for $x > 0$, $x \neq 1$.

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

$\frac{dy}{dx} = \frac{1 \cdot \ln(x) - x \cdot \frac{1}{x}}{[\ln(x)]^2} = \frac{\ln(x) - 1}{[\ln(x)]^2}$.

The chain rule

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

In words: differentiate the outside leaving the inside alone, then multiply by the derivative of the inside.

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

Inside: $u = 3 x^2$, $\frac{du}{dx} = 6 x$. Outside: $\frac{dy}{du} = \cos(u)$.

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

Chain rule shortcuts for common composites

These are worth memorising as patterns:

• IMATH_60
• IMATH_61 for $f(x) > 0$
• IMATH_63
• IMATH_64
• IMATH_65

Combining rules

Many Paper 1 questions combine two or three rules.

Worked example. Differentiate $f(x) = x^2 e^{3 x}$.

Product rule with $u = x^2$, $v = e^{3 x}$. $u' = 2 x$, $v' = 3 e^{3 x}$ (chain rule on the inside).

$f'(x) = 2 x \cdot e^{3 x} + x^2 \cdot 3 e^{3 x} = e^{3 x} (2 x + 3 x^2) = x e^{3 x} (2 + 3 x)$.

Worked example. Differentiate $f(x) = \ln(\cos(x))$.

Chain rule with outside $\ln$ and inside $\cos(x)$.

$f'(x) = \frac{1}{\cos(x)} \cdot (-\sin(x)) = -\frac{\sin(x)}{\cos(x)} = -\tan(x)$.

Common Paper 1 traps

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

Sign error in the quotient rule. The numerator is $u' v$ minus $u v'$, in that order. Reversing it flips the sign.

Power rule on $a^x$ for $a \neq e$. $\frac{d}{dx}(2^x)$ is not $x \cdot 2^{x - 1}$. Use $2^x = e^{x \ln 2}$ and the chain rule: $\frac{d}{dx}(2^x) = (\ln 2) \cdot 2^x$.

Treating $e^{2x}$ as $2 e^{2x}$ without the chain rule. The chain rule gives $\frac{d}{dx}(e^{2x}) = 2 e^{2x}$, but the working should show the chain rule explicitly.

Stopping before simplification. Markers usually reward a clean factored form. After the quotient rule, look for common factors in numerator and denominator.

In one sentence

The four differentiation rules (sum, product, quotient, chain) combined with the standard derivatives of $x^n$, $e^x$, $\ln(x)$, $\sin(x)$, $\cos(x)$ and $\tan(x)$ let you differentiate any function in the VCE Math Methods Paper 1 library, with the chain rule the single most-tested rule and the product and quotient rules requiring careful attention to signs.