Differentiating trigonometric functions

The derivatives of sin x, cos x and tan x in radians, the chain rule standard forms, and differentiating products, quotients, powers and compositions involving trigonometric functions.

Standard forms

Derivatives (x in radians)

$$\frac{d}{dx}\sin x = \cos x \qquad \frac{d}{dx}\cos x = -\sin x \qquad \frac{d}{dx}\tan x = \sec^2 x$$

The gradient of sin x is cos x

Drag the point along y = sin x. The gradient readout always matches the height of the dashed cosine curve.

Chain rule forms

$$\frac{d}{dx}\sin f(x) = f'(x)\cos f(x) \qquad \frac{d}{dx}\cos f(x) = -f'(x)\sin f(x) \qquad \frac{d}{dx}\tan f(x) = f'(x)\sec^2 f(x)$$

Example 1●

Differentiate (a) $\sin 3x$, (b) $\cos(2x - 1)$ and (c) $\tan 5x$.

$3\cos 3x$

$-2\sin(2x - 1)$

$5\sec^2 5x$

Example 2●●

Differentiate (a) $\sin x^2$ and (b) $\cos^3 x$.

$2x\cos x^2$

$\cos^3 x = (\cos x)^3$, so the derivative is $3\cos^2 x \times (-\sin x) = -3\cos^2 x\sin x$.

Products, quotients and other functions

Example 3●●

Differentiate (a) $x\sin x$, (b) $\dfrac{\sin x}{x}$ and (c) $\ln(\cos x)$.

$\sin x + x\cos x$

$\dfrac{x\cos x - \sin x}{x^2}$

$\dfrac{-\sin x}{\cos x} = -\tan x$

Flashcards

$dfrac{d}{dx}sin ax$

$a\cos ax$

$dfrac{d}{dx}cos ax$

$-a\sin ax$

$dfrac{d}{dx} an ax$

$a\sec^2 ax$

Practice

$\dfrac{d}{dx}\sin 3x = $
$\dfrac{d}{dx}\cos(2x - 1) = $
$\dfrac{d}{dx}\tan 5x = $
$\dfrac{d}{dx}\sin x^2 = $
$\dfrac{d}{dx}\cos^3 x = $
$\dfrac{d}{dx}x\sin x = $
$\dfrac{d}{dx}\ln(\cos x) = $
Find the gradient of $y = \sin 2x$ at $x = 0$.
Find the gradient of $y = \tan x$ at $x = \dfrac{\pi}{4}$.