Derivatives of trigonometric functions: WACE Year 12 Mathematics Methods Unit 3

What this dot point is asking

SCSA Unit 3 requires fluency differentiating the three core trigonometric functions and their composites. These derivatives appear in both the calculator-free and calculator-assumed sections and combine constantly with the product, quotient and chain rules.

The standard derivatives

The derivatives below assume the angle is measured in radians. The simple form $\dfrac{d}{dx}\sin x=\cos x$ only holds because $\lim_{h\to0}\dfrac{\sin h}{h}=1$ when $h$ is in radians; in degrees an awkward constant factor appears.

Note the sign: differentiating cosine introduces a minus, while differentiating sine does not. The derivative of $\tan x$ follows from the quotient rule applied to $\dfrac{\sin x}{\cos x}$.

The chain rule with trigonometric functions

When the angle is a function of $x$, the chain rule multiplies by the derivative of that angle. For $y=\sin(kx)$ the angle $kx$ has derivative $k$, so $y'=k\cos(kx)$.

Combining with other rules

Trigonometric derivatives frequently sit inside a product or quotient. The strategy is to label each factor and differentiate it, applying the chain rule to any composite angle.

The derivative of $\tan x$

The tangent derivative follows from the quotient rule on $\tan x=\dfrac{\sin x}{\cos x}$:

The Pythagorean identity $\cos^{2}x+\sin^{2}x=1$ collapses the numerator to $1$. With the chain rule, $\dfrac{d}{dx}\tan(f(x))=f'(x)\sec^{2}(f(x))$.

Stationary points of trigonometric curves

Differentiating trig functions lets you locate the turning points of oscillating models. For $y=\sin x$, $y'=\cos x=0$ at $x=\dfrac{\pi}{2},\dfrac{3\pi}{2},\dots$, which are exactly the peaks and troughs. Because the second derivative of $\sin x$ is $-\sin x$, a peak at $x=\dfrac{\pi}{2}$ has $y''=-1<0$ (a maximum), confirming the shape. This connects directly to motion problems where velocity is the derivative of displacement.

Higher derivatives cycle through four functions

Repeatedly differentiating $\sin x$ produces a cycle of period four: $\sin x\to\cos x\to-\sin x\to-\cos x\to\sin x$. So the fourth derivative of $\sin x$ returns to $\sin x$. This pattern is useful for finding the second derivative quickly when classifying stationary points of trigonometric models, and it explains why simple harmonic motion satisfies $\dfrac{d^{2}x}{dt^{2}}=-\omega^{2}x$.

Why radians are required

All these derivative rules are only valid when $x$ is in radians. If a question is set in degrees you must convert, or differentiate $\sin\!\left(\dfrac{\pi x}{180}\right)$ with the chain rule, which produces the awkward factor $\dfrac{\pi}{180}$. SCSA exam questions involving calculus use radians.