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Pure · Differentiation

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Differentiating trigonometric functions

The six standard trig derivatives — sin→cos, cos→−sin, tan→sec², sec→sec tan, cosec→−cosec cot, cot→−cosec² — why x must be in radians, and combining them with the chain, product and quotient rules.

Pure · Differentiation

Differentiating trigonometric functions

The six standard trig derivatives — sin→cos, cos→−sin, tan→sec², sec→sec tan, cosec→−cosec cot, cot→−cosec² — why x must be in radians, and combining them with the chain, product and quotient rules.

Why it works

Start with the most important one, from first principles. The derivative of sin⁡x\sin x is the limit of sin⁡(x+h)−sin⁡xh\dfrac{\sin(x+h) - \sin x}{h} as h→0h \to 0. Expand the top with the addition formula sin⁡(x+h)=sin⁡xcos⁡h+cos⁡xsin⁡h\sin(x+h) = \sin x\cos h + \cos x\sin h: sin⁡xcos⁡h+cos⁡xsin⁡h−sin⁡xh=sin⁡x⋅cos⁡h−1h+cos⁡x⋅sin⁡hh.\frac{\sin x\cos h + \cos x\sin h - \sin x}{h} = \sin x\cdot\frac{\cos h - 1}{h} + \cos x\cdot\frac{\sin h}{h}. Now use the small-angle results for tiny hh (in radians): sin⁡h≈h\sin h \approx h, so sin⁡hh→1\frac{\sin h}{h}\to 1; and cos⁡h≈1−12h2\cos h \approx 1 - \tfrac12 h^2, so cos⁡h−1h≈−12h→0\frac{\cos h - 1}{h}\approx -\tfrac12 h \to 0. The first term vanishes and the second leaves cos⁡x\cos x: ddxsin⁡x=cos⁡x.\frac{d}{dx}\sin x = \cos x. The same calculation on cos⁡x\cos x gives ddxcos⁡x=−sin⁡x\dfrac{d}{dx}\cos x = -\sin x (note the minus — cosine is falling where sine is positive).

This only works in radians. The step sin⁡hh→1\frac{\sin h}{h}\to 1 is true for hh in radians and false in degrees (there it tends to π180\frac{\pi}{180}). So every trig derivative on this page assumes xx is in radians — differentiate sin⁡(x∘)\sin(x^\circ) and you pick up an unwanted factor of π180\frac{\pi}{180}. In calculus, angles are always radians.

The other four come from these two using the quotient rule. For example tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{\sin x}{\cos x} gives ddxtan⁡x=cos⁡xcos⁡x−sin⁡x(−sin⁡x)cos⁡2x=cos⁡2x+sin⁡2xcos⁡2x=1cos⁡2x=sec⁡2x.\frac{d}{dx}\tan x = \frac{\cos x\cos x - \sin x(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x. Differentiating sec⁡x=(cos⁡x)−1\sec x = (\cos x)^{-1}, cosec⁡x=(sin⁡x)−1\operatorname{cosec} x = (\sin x)^{-1} and cot⁡x=cos⁡xsin⁡x\cot x = \frac{\cos x}{\sin x} the same way gives the full set worth memorising: ddxsin⁡x=cos⁡x,ddxcos⁡x=−sin⁡x,ddxtan⁡x=sec⁡2x,\frac{d}{dx}\sin x = \cos x, \qquad \frac{d}{dx}\cos x = -\sin x, \qquad \frac{d}{dx}\tan x = \sec^2 x, ddxsec⁡x=sec⁡xtan⁡x,ddxcosec⁡x=−cosec⁡xcot⁡x,ddxcot⁡x=−cosec⁡2x.\frac{d}{dx}\sec x = \sec x\tan x, \quad \frac{d}{dx}\operatorname{cosec} x = -\operatorname{cosec} x\cot x, \quad \frac{d}{dx}\cot x = -\operatorname{cosec}^2 x. The three "co-" functions all carry a minus sign — a handy way to remember which is which.

With the chain rule these extend to any inside function: ddxsin⁡(kx)=kcos⁡(kx)\frac{d}{dx}\sin(kx) = k\cos(kx), and ddxsin⁡(f(x))=cos⁡(f(x)) f′(x)\frac{d}{dx}\sin(f(x)) = \cos(f(x))\,f'(x). A very common slip is sin⁡2x\sin^2 x: that means (sin⁡x)2(\sin x)^2, so it is the chain rule on a square — ddxsin⁡2x=2sin⁡xcos⁡x\frac{d}{dx}\sin^2 x = 2\sin x\cos x, not cos⁡2x\cos^2 x.