AS 91578 · 2025 paper · derivatives, normal, tangents

Question 1(c) Merit

Practises The rules & standard derivatives

A chain-rule power, an oven cooling written in half-powers of t, a trig normal, a tangent to an exponential-plus-quadratic, and a parametric curve where inflection and stationary have to coincide.

What is the gradient of the normal to the curve y = sin³ x cos x when x = π/4?

The 2025 paper, page 2 ↗ Marking schedule, page 1 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1 Is this a product, a function inside a function, or both?
Both. It is sin³ x times cos x, and sin³ x = (sin x)³ needs the chain rule.
2 Differentiate each factor on its own.
u = sin³ x, so u' = 3 sin² x cos x
v = cos x, so v' = −sin x
3 Put them together with the product rule.
dy/dx = 3 sin² x cos x · cos x + sin³ x · (−sin x) = 3 sin² x cos² x − sin⁴ x
4 At x = π/4, what are sin x and cos x? Find the curve’s gradient.
Both are 1/√2, so sin² x = cos² x = 1/2 and sin⁴ x = 1/4.
dy/dx = 3 × 1/2 × 1/2 − 1/4 = 3/4 − 1/4 = 1/2
5 The question wants the normal, not the tangent. How are their gradients related?
The normal is at right angles to the tangent, so its gradient is −1 ÷ 1/2 = −2.

Answer The tangent’s gradient is 1/2, so the normal’s gradient is −2.

What the marker wanted

Achieved The correct derivative.
Merit Goes on to the normal: a gradient of −2, worked out from that derivative.

In our words, from NZQA’s 2025 marking schedule.

Common mistake

Losing a factor while differentiating, usually the cos x that the chain rule brings out of sin³ x. NZQA’s 2025 report says many students did, and a wrong derivative here earned nothing, even when everything after it was right. Others tried to rewrite it with trig identities first and went wrong that way.

How did it go?

Got it moves it further out. Review brings it back tomorrow.