Quick questions on Derivatives of trigonometric functions in radians (QCE Mathematical Methods Unit 3)

In Methods, all calculus on trig functions is done with $x$ in radians. The formula $\frac{d}{dx}(\sin x) = \cos x$ is only true when $x$ is in radians. If you work in degrees the derivative picks up an awkward factor of $\frac{\pi}{180}$, which is why QCAA requires radians for calculus questions and why most calculators default to degree mode on a fresh reset (always check and switch to radians).

$$\frac{d}{dx}(\sin x) = \cos x$$

For any differentiable inner $f(x)$:

$$\frac{d}{dx}\bigl(\sin(a x + b)\bigr) = a \cos(a x + b)$$

Differentiate $y = \cos(3x - 1)$.

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

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

A particle moves so that its displacement from the origin is $s(t) = 4 \sin(2 t)$ metres, with $t$ in seconds. Find its velocity at $t = \dfrac{\pi}{6}$.

$\dfrac{dy}{dx} = 3 \sec^2(3x) = \dfrac{3}{\cos^2(3x)}$.

$\frac{d}{dx}(\sin x) = \cos x$ only when $x$ is in radians. Set your calculator to radians for Paper 2. Paper 1 problems are always in radians by convention.

$\frac{d}{dx}(\cos x) = -\sin x$. Reversing the sign turns a maximum into a minimum and a velocity into its negative.

$\frac{d}{dx}(\sin(2x)) = 2 \cos(2x)$, not $\cos(2x)$. The factor of $2$ comes from differentiating the inner $2x$.

Paper 1 expects $\sin(\pi/6) = 1/2$, $\cos(\pi/4) = \sqrt{2}/2$, $\tan(\pi/3) = \sqrt{3}$ and the related angle values without a calculator. Drill the unit circle.

$\sin^2 x + \cos^2 x = 1$, not $\sin^2 x - \cos^2 x = 1$. Sign errors here lead to wrong derivatives on optimisation questions.