Further Differentiation of Trigonometric Functions

Video lesson · from 1:51:474 worked examples

One lesson video covers all of Differentiation, so it opens at 1:51:47 for this topic — not from the beginning.

The chain rule is also required when differentiating composite trigonometric functions.

\frac{d}{dx}[\sin(f(x))] = \cos(f(x)) \times f'(x)

\frac{d}{dx}[\cos(f(x))] = -\sin(f(x)) \times f'(x)

Two things must happen: differentiate the trig function and multiply by the derivative of the angle. $\sin(4x)$ differentiates to $4\cos(4x)$ .
The minus still belongs to cosine: $\cos(3x)$ differentiates to $-3\sin(3x)$ — both the chain factor and the minus sign are needed.
Powers of trig functions are chains too: $\sin^3 x$ means $(\sin x)^3$ , so it differentiates to $3\sin^2 x \cos x$ .
Work in radians: the derivatives of $\sin$ and $\cos$ are only valid in radians.

Example 1

If $y = \sin 4x$ , find $\frac{dy}{dx}$ .

Apply the chain rule. The derivative of $\sin u$ is $\cos u$ .

\begin{aligned} \frac{dy}{dx} &= \cos 4x \times \frac{d}{dx}(4x) \\ &= \cos 4x \times 4 \\ &= 4\cos 4x \end{aligned}

Example 2

If $y = \sin(4x + 3)$ , find $\frac{dy}{dx}$ .

\begin{aligned} \frac{dy}{dx} &= \cos(4x + 3) \times \frac{d}{dx}(4x + 3) \\ &= \cos(4x + 3) \times 4 \\ &= 4\cos(4x + 3) \end{aligned}

Example 3

If $y = \cos(x^2)$ , find $\frac{dy}{dx}$ .

\begin{aligned} \frac{dy}{dx} &= -\sin(x^2) \times \frac{d}{dx}(x^2) \\ &= -\sin(x^2) \times 2x \\ &= -2x\sin(x^2) \end{aligned}

Example 4

If $y = \sin^3 x$ , find $\frac{dy}{dx}$ .

First, rewrite the function to clearly see the composite structure:

y = (\sin x)^3

Apply the chain rule, treating $\sin x$ as the inner bracket:

\begin{aligned} \frac{dy}{dx} &= 3(\sin x)^2 \times \frac{d}{dx}(\sin x) \\ &= 3(\sin x)^2 \times \cos x \\ &= 3\sin^2 x \cos x \end{aligned}