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VICMath MethodsSyllabus dot point

How are trigonometric functions differentiated?

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

A focused answer to the VCE Maths Methods Unit 2 dot point on derivatives of trig functions. States ddxsinx=cosx\frac{d}{dx} \sin x = \cos x, ddxcosx=sinx\frac{d}{dx} \cos x = -\sin x, ddxtanx=sec2x\frac{d}{dx} \tan x = \sec^2 x, the chain-rule extensions, and works the VCAA SAC-style oscillation-derivative problem.

Generated by Claude Opus 4.87 min answer

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  1. What this dot point is asking
  2. Standard derivatives
  3. Chain rule extensions
  4. Standard combinations
  5. Connection to oscillation
  6. Choosing the right rule
  7. Finding tangents and stationary points
  8. In one sentence
  9. Examples in context
  10. Try this

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

ddxsinx=cosx\frac{d}{dx} \sin x = \cos x

ddxcosx=sinx\frac{d}{dx} \cos x = -\sin x

ddxtanx=sec2x=1cos2x\frac{d}{dx} \tan x = \sec^2 x = \frac{1}{\cos^2 x}

These derivations use the small-angle limit limθ0sinθ/θ=1\lim_{\theta \to 0} \sin\theta/\theta = 1 together with the addition formulas. The cosine result picks up the minus sign because cosθ\cos\theta decreases as θ\theta moves from zero.

Chain rule extensions

ddxsin(u(x))=u(x)cos(u(x))\frac{d}{dx} \sin(u(x)) = u'(x) \cos(u(x))

ddxcos(u(x))=u(x)sin(u(x))\frac{d}{dx} \cos(u(x)) = -u'(x) \sin(u(x))

ddxtan(u(x))=u(x)sec2(u(x))\frac{d}{dx} \tan(u(x)) = u'(x) \sec^2(u(x))

Standard combinations

Trig times polynomial
Product rule. ddx(xsinx)=sinx+xcosx\frac{d}{dx} (x \sin x) = \sin x + x \cos x.
Squared trig
Chain rule with u2u^2. ddxsin2x=2sinxcosx=sin2x\frac{d}{dx} \sin^2 x = 2 \sin x \cos x = \sin 2x.
Trig over polynomial
Quotient rule.

Connection to oscillation

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

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

a(t)=dv/dt=Aω2sin(ωt)=ω2x(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.

Choosing the right rule

Trig derivatives appear inside products, quotients and composites. A product such as xsinxx\sin x uses the product rule; a power such as sin2x=(sinx)2\sin^2 x = (\sin x)^2 uses the chain rule with outer power u2u^2; a quotient such as sinxx\dfrac{\sin x}{x} uses the quotient rule; and a nested argument such as sin(x2+1)\sin(x^2 + 1) uses the chain rule on the argument. When more than one rule applies, work from the outside in: identify the outermost operation, apply its rule, and differentiate the inner part as a sub-problem. Many VCAA items then ask you to set the derivative to zero, so simplifying with double-angle identities (for example 2sinxcosx=sin2x2\sin x\cos x = \sin 2x) often makes the equation solvable by hand.

Finding tangents and stationary points

The derivative gives the gradient of the tangent at any point, so for y=f(x)y = f(x) the tangent at x=ax = a has gradient f(a)f'(a) and equation yf(a)=f(a)(xa)y - f(a) = f'(a)(x - a). Stationary points occur where f(x)=0f'(x) = 0; for trig functions these recur periodically, so always restrict to the stated interval and report every solution in it. Because sin\sin and cos\cos oscillate, a trig function typically has alternating maxima and minima, which a sign test on ff' confirms.

In one sentence

The standard derivatives are ddxsinx=cosx\frac{d}{dx} \sin x = \cos x, ddxcosx=sinx\frac{d}{dx} \cos x = -\sin x, ddxtanx=sec2x\frac{d}{dx} \tan x = \sec^2 x, with chain-rule extensions ddxsinu=ucosu\frac{d}{dx} \sin u = u' \cos u and equivalents, assuming xx is in radians throughout.

Examples in context

Example 1. Velocity of an oscillating piston. A piston's displacement is x(t)=4sin(2t)x(t) = 4\sin(2t) cm, with tt in seconds. Its velocity is x(t)=4×2cos(2t)=8cos(2t)x'(t) = 4 \times 2\cos(2t) = 8\cos(2t) cm/s. The maximum speed is 88 cm/s, occurring whenever cos(2t)=±1\cos(2t) = \pm 1, that is at t=0,π2,π,t = 0, \frac{\pi}{2}, \pi, \dots (the centre of each swing).

Example 2. Slope of a tidal curve. A tide height is h(t)=3cos ⁣(π6t)h(t) = 3\cos\!\left(\frac{\pi}{6}t\right) m, tt in hours. The rate of change is h(t)=3×π6sin ⁣(π6t)=π2sin ⁣(π6t)h'(t) = -3 \times \frac{\pi}{6}\sin\!\left(\frac{\pi}{6}t\right) = -\frac{\pi}{2}\sin\!\left(\frac{\pi}{6}t\right) m/h. At t=3t = 3 (a quarter period), h(3)=π2sin ⁣(π2)=π21.57h'(3) = -\frac{\pi}{2}\sin\!\left(\frac{\pi}{2}\right) = -\frac{\pi}{2} \approx -1.57 m/h, the fastest fall.

Try this

Q1. Differentiate f(x)=cos(5x)f(x) = \cos(5x) and g(x)=tan(3x)g(x) = \tan(3x). [2+2 marks]

  • Cue. f(x)=5sin(5x)f'(x) = -5\sin(5x); g(x)=3sec2(3x)g'(x) = 3\sec^2(3x).

Q2. Find the derivative of y=xcosxy = x\cos x. [3 marks]

  • Cue. Product rule: y=cosxxsinxy' = \cos x - x\sin x.

Q3. A particle's displacement is s(t)=2sin(3t)s(t) = 2\sin(3t) m. (a) Find the velocity v(t)v(t). (b) Find the velocity at t=0t = 0. [2+2 marks]

  • Cue. (a) v(t)=6cos(3t)v(t) = 6\cos(3t). (b) v(0)=6v(0) = 6 m/s.

Exam-style practice questions

Practice questions written in the style of VCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

VCAA 2022 Exam 14 marksDifferentiate (a) f(x)=sin(3x)f(x) = \sin(3x) and (b) g(x)=cos2(x)g(x) = \cos^2(x).
Show worked answer →

(a) Chain rule on sinu\sin u. u=3xu = 3x, du/dx=3du/dx = 3.

f(x)=cos(3x)3=3cos(3x)f'(x) = \cos(3x) \cdot 3 = 3\cos(3x).

(b) Chain rule on u2u^2 with u=cosxu = \cos x. du/dx=sinxdu/dx = -\sin x.

g(x)=2cosx(sinx)=2sinxcosx=sin(2x)g'(x) = 2 \cos x \cdot (-\sin x) = -2 \sin x \cos x = -\sin(2x) (using the double-angle identity).

Markers reward chain rule, sign on sinx-\sin x, and the optional double-angle simplification.

VCAA 2023 Exam 25 marksLet f(x)=sinx+3cosxf(x) = \sin x + \sqrt{3}\cos x for x[0,2π]x \in [0, 2\pi]. (a) Find f(x)f'(x). (b) Hence determine the xx-coordinates of the stationary points of ff on the interval.
Show worked answer →

(a) Differentiating term by term, f(x)=cosx3sinxf'(x) = \cos x - \sqrt{3}\sin x.

(b) Stationary points where f(x)=0f'(x) = 0: cosx3sinx=0\cos x - \sqrt{3}\sin x = 0, so cosx=3sinx\cos x = \sqrt{3}\sin x, giving tanx=13\tan x = \dfrac{1}{\sqrt{3}}.

The reference angle is π6\dfrac{\pi}{6}, and tanx>0\tan x > 0 in quadrants 1 and 3, so over [0,2π][0, 2\pi], x=π6x = \dfrac{\pi}{6} and x=π6+π=7π6x = \dfrac{\pi}{6} + \pi = \dfrac{7\pi}{6}.

Markers reward differentiating both terms, setting f(x)=0f'(x) = 0, reducing to tanx=13\tan x = \dfrac{1}{\sqrt{3}}, and giving both solutions in the interval.

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