Differentiate sine, cosine and tangent functions and their compositions via the chain rule

What this dot point is asking

VCAA wants you to differentiate sine, cosine and tangent functions and their composites using the chain rule.

Standard derivatives

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These derivations use the small-angle limit $\lim_{\theta \to 0} \sin\theta/\theta = 1$ together with the addition formulas. The cosine result picks up the minus sign because $\cos\theta$ decreases as $\theta$ moves from zero.

Chain rule extensions

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Standard combinations

Trig times polynomial. Product rule. $\frac{d}{dx} (x \sin x) = \sin x + x \cos x$.

Squared trig. Chain rule with $u^2$. $\frac{d}{dx} \sin^2 x = 2 \sin x \cos x = \sin 2x$.

Trig over polynomial. Quotient rule.

Connection to oscillation

If $x(t) = A \sin(\omega t)$ is the displacement of a simple oscillator, then:

$v(t) = dx/dt = A \omega \cos(\omega t)$

$a(t) = dv/dt = -A \omega^2 \sin(\omega t) = -\omega^2 x(t)$

This is the differential equation of simple harmonic motion. The derivative pattern explains why oscillators have constant period independent of amplitude.

Worked example

Differentiate $h(x) = \sin(x^2 + 1)$.

Chain rule: $u = x^2 + 1$, $du/dx = 2x$.

$h'(x) = \cos(x^2 + 1) \cdot 2x = 2x \cos(x^2 + 1)$.

Common traps

Sign on cosine derivative. $\frac{d}{dx} \cos x = -\sin x$. The minus is the most-common slip.

Forgetting the chain factor. $\frac{d}{dx} \sin(3x) = 3\cos(3x)$, not $\cos(3x)$.

Confusing $\sin^2 x$ with $\sin(x^2)$. $\sin^2 x = (\sin x)^2$. $\sin(x^2)$ is the sine of $x^2$. Different functions, different derivatives.

Working in degrees. Calculus assumes radians throughout. $\sin x$ has derivative $\cos x$ only when $x$ is in radians.

In one sentence

The standard derivatives are $\frac{d}{dx} \sin x = \cos x$, $\frac{d}{dx} \cos x = -\sin x$, $\frac{d}{dx} \tan x = \sec^2 x$, with chain-rule extensions $\frac{d}{dx} \sin u = u' \cos u$ and equivalents, assuming $x$ is in radians throughout.