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 $\tan$.

What the derivative means

Every differentiation rule is a tool for finding $f'(x)$, the instantaneous rate of change of $f$, which is also the gradient of the tangent to $y = f(x)$ at each point. The first-principles definition behind every rule is

NESA can ask you to differentiate from first principles for a simple polynomial, so you should be able to set up this limit, expand $f(x + h)$, cancel the $h$ in the numerator, and take the limit. For all other functions you quote the standard rules above, which are themselves derived from this limit.

Seeing the derivative: from secant to tangent

The derivative is the gradient of the tangent, and the tangent is the limiting position of a secant. Watching that limit form is the single best way to understand what $f'(x)$ measures. Fix a point $P$ on the curve $y = f(x)$ and let a second point $Q$ slide towards it; the gradient of the line $PQ$ is the average rate of change $\frac{f(a+h) - f(a)}{h}$, and as $h \to 0$ it tends to the instantaneous rate of change at $P$.

Stage 1, a wide secant. Place a fixed point $P$ at $x = a$ and a second point $Q$ a large horizontal distance $h$ to the right. The straight line through $P$ and $Q$ is a secant, and its gradient is the average rate of change between the two points, the rise divided by the run.

Stage 2, slide $Q$ closer. Move $Q$ towards $P$ so the gap $h$ shrinks. The secant now hugs the curve more closely near $P$, and its gradient is a better estimate of the gradient at $P$ itself.

Stage 3, let $h \to 0$. As $Q$ slides almost on top of $P$, the gap $h$ becomes tiny and the secant is barely distinguishable from a line that just grazes the curve at $P$. The average rate of change is converging on a single number.

Stage 4, the tangent is the limit. In the limit $h \to 0$, $Q$ merges with $P$ and the secant becomes the tangent at $P$. Its gradient is the derivative $f'(a)$, the instantaneous rate of change. Every rule on this page is a shortcut for computing exactly this limiting gradient.

Deciding which rule to use

The single most common student error is reaching for the wrong rule. Read the structure of the expression first.

• A sum or difference of terms: differentiate each term separately.
• A function raised to a power, or a function inside another function (a composition): chain rule.
• Two functions multiplied together: product rule.
• One function divided by another: quotient rule.

Many HSC expressions combine rules. For $y = x^2 \sin(2x)$ you use the product rule on the outer multiplication and the chain rule on $\sin(2x)$. Work from the outside in, naming each piece as you go.

Tangents and normals

Once you have $f'(x)$, the gradient of the tangent at $x = a$ is $m = f'(a)$. The tangent line is

The normal is perpendicular to the tangent, so its gradient is $-\frac{1}{f'(a)}$ (provided $f'(a) \neq 0$). These appear constantly in HSC questions: differentiate, evaluate the derivative at the point, then write the line equation.

Higher derivatives

Differentiating $f'(x)$ again gives the second derivative $f''(x)$, the rate of change of the gradient. For $f(x) = x^3$, $f'(x) = 3x^2$ and $f''(x) = 6x$. Higher derivatives drive concavity and motion problems, so the rules here feed directly into later calculus topics.

How exam questions ask about differentiation

The wording varies, but each version is a cue for a particular rule or follow-up:

• "Differentiate" or "find $\frac{dy}{dx}$". A direct rule application. Identify the structure (sum, composition, product, quotient) first, then apply the matching rule and simplify.
• "Differentiate from first principles". Set up $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$, expand $f(x+h)$, cancel the $h$, and take the limit. NESA reserves this for simple polynomials, but you must show the limit, not quote the power rule.
• "Find the gradient of the curve at $x = a$". Differentiate, then substitute $x = a$ into $f'(x)$. The number you get is the gradient of the tangent there.
• "Find the equation of the tangent / normal at the point ...". Differentiate, evaluate $f'(a)$ for the gradient, then use $y - f(a) = f'(a)(x - a)$ for the tangent; for the normal use gradient $-\frac{1}{f'(a)}$.
• "Show that $\frac{dy}{dx} = \ldots$". A "show that" answer must reach the printed result by valid steps; do not just assert it. Pick up the marks by displaying the rule used and every simplification.
• A derivative buried in a larger question. Most calculus questions open with a differentiation step that feeds a later part (stationary points, rates, areas). Get it right and factored, because every later mark depends on it.