How to Differentiate Trigonometric Functions. Visual Explanation with color coded examples

# How to Differentiate Trigonometric Functions

### Quick Overview

• $$\displaystyle \frac d {dx}\left(\sin kx\right) = k\cos kx$$
• $$\displaystyle \frac d {dx}\left(\cos kx\right) = -k\sin kx$$
• $$\displaystyle \frac d {dx}\left(\tan kx\right) = k\sec^2 kx$$
• $$\displaystyle \frac d {dx}\left(\cot kx\right) = -k\csc^2 kx$$
• $$\displaystyle \frac d {dx}\left(\sec kx\right) = k\sec kx\tan kx$$
• $$\displaystyle \frac d {dx}\left(\csc kx\right) = -k\csc kx\cot kx$$
• Notice that the derivatives of the co-functions are negative. That is, the derivative of the cosine, cotangent, and cosecant are the ones with negative signs.
• The trig functions are paired when it comes to differentiation: sine and cosine, tangent and secant, cotangent and cosecant.
• This lesson assumes you are familiar with the Power Rule, Product Rule, Quotient Rule and Chain Rule.

### Derivations of the Derivatives of Trig Functions

The derivatives of each of the trig functions was derived in a previous lesson. If you would like to see why the derivatives are what they are, here are links to the lessons where the derivations are given:

Derivatives of the sine and cosine:
Derivatives of the tangent and cotangent:
Derivatives of the secant and cosecant:

It may seem overwhelming to try and remember all of these derivatives, but there are three patterns that can be helpful.

Pattern 1: Negatives and the Co-functions.

Only the derivatives of the co-functions have a negative sign.

$$\begin{array}{lcl} \frac d {dx}\left(\sin x\right) = \cos x & \hspace{15mm} & \frac d {dx}\left(\cos x\right) = -\sin x\\[6pt] % \frac d {dx}\left(\tan x\right) = \sec^2 x & & \frac d {dx}\left(\cot x\right) = -\csc^2 x\\[6pt] % \frac d {dx}\left(\sec x\right) = \sec x\tan x & & \frac d {dx}\left(\csc x\right) = -\csc x\cot x\\ \end{array}$$

Pattern 2: Pairs of Functions

The trig functions are paired when it comes to derivatives.

$$\frac d {dx}\left(\sin x\right) = \cos x \\ \frac d {dx} \left(\cos x\right) = -\sin x \\ \hspace{1mm} \\ \frac d {dx}\left(\tan x\right) = \sec^2 x \\ \frac d {dx}\left(\sec x\right) = \sec x \tan x \\ \hspace{1mm} \\ \frac d {dx}\left(\cot x\right) = -\csc^2 x \\ \frac d {dx}\left(\csc x\right) = -\csc x\cot x$$ Pattern 3: Each Type of Trig Function has its own Pattern

Sines and Cosines have Single Function Derivatives

$$\frac d {dx}\left(\sin x\right) = \cos x \\ \frac d {dx} \left(\cos x\right) = -\sin x$$

Tangent and Cotangents have Squared Derivatives

$$\frac d {dx}\left(\tan x\right) = \sec^2 x \\ \frac d {dx}\left(\cot x\right) = -\csc^2 x$$

Secants and Cosecants have Two Function Derivatives

$$\frac d {dx}\left(\sec x\right) = \sec x \tan x \\ \frac d {dx}\left(\csc x\right) = -\csc x\cot x$$

### Examples

##### Example 1

Find $$\displaystyle \frac d {dx}\left(\cos(x^2+9)\right)$$.

Step 1

Use the chain rule to differentiate.

$$\frac d {dx}\left(\cos(x^2+9)\right) = -\sin(x^2+9)\cdot \frac d {dx}\left(x^2 + 9\right) = -\sin(x^2+9)\cdot 2x = -2x\sin(x^2+9)$$

$$\displaystyle \frac d {dx}\left(\cos(x^2+9)\right) = -2x\sin(x^2+9)$$

##### Example 2

Suppose $$f(x) = x^4\tan 3x$$. Find $$f'(x)$$.

Step 1

Identify the factors in the function.

$$f(x) = \blue{x^4}\red{\tan 3x}$$

Step 2

Differentiate using the product rule.

\begin{align*}% f'(x) & = \blue{4x^3}\tan 3x + x^4\cdot \red{3\sec^2 3x}\\ & = 4x^3\tan 3x + 3x^4\sec^2 3x \end{align*}

Step 3

Simplify by factoring.

\begin{align*} f'(x) & = 4\blue{x^3}\tan 3x + 3\blue{x^4}\sec^2 3x\\ & = \blue{x^3}\left(4\tan 3x + 3x\sec^2 3x\right) \end{align*}

$$\displaystyle f'(x) = x^3\left(4\tan 3x + 3x\sec^2 3x\right)$$.

##### Example 3

Find $$\displaystyle \frac d {dx}\left(\frac{\sin 8x}{1 + \sec 8x}\right)$$.

Step 1

Differentiate using the quotient rule. The parts in $$\blue{blue}$$ are related to the numerator.

\begin{align*} \frac d {dx}\left(\frac{\sin 8x}{1 + \sec 8x}\right) & = \frac{(1 + \sec 8x)\cdot\blue{8\cos 8x} - \blue{\sin 8x}\cdot 8\sec 8x\tan 8x}{(1 + \sec 8x)^2} \end{align*}

Step 2

Simplify the numerator.

\begin{align*} \frac d {dx}\left(\frac{\sin 8x}{1 + \sec 8x}\right) & = \frac{(1 + \sec 8x)\cdot \blue 8\cos 8x - \sin 8x\cdot \blue 8\sec 8x\tan 8x}{(1 + \sec 8x)^2}\\[6pt] & = \frac{\blue 8(1 + \red{\sec 8x})\cos 8x - \blue 8\sin 8x\red{\sec 8x}\tan 8x}{(1 + \sec 8x)^2}\\[6pt] & = \frac{8\left(1 + \red{\frac 1 {\cos 8x}}\right)\cos 8x - 8\sin 8x\cdot\red{\frac 1 {\cos 8x}}\cdot\tan 8x}{(1 + \sec 8x)^2}\\[6pt] & = \frac{8\left(\cos 8x + \red 1\right) - 8\cdot \frac{\sin 8x}{\red{\cos 8x}}\cdot\tan 8x}{(1 + \sec 8x)^2}\\[6pt] & = \frac{8\left(\cos 8x + 1\right) - 8\tan 8x\tan 8x}{(1 + \sec 8x)^2}\\[6pt] & = \frac{8\left(\cos 8x + 1\right) - 8\tan^2 8x}{(1 + \sec 8x)^2}\\[6pt] & = \frac{8\left(\cos 8x + 1 - \tan^2 8x\right)}{(1 + \sec 8x)^2}\\[6pt] & = \frac{8\left(\cos 8x - \tan^2 8x + 1\right)}{(1 + \sec 8x)^2} \end{align*}

$$\displaystyle \frac d {dx}\left(\frac{\sin 8x}{1 + \sec 8x}\right) = \frac{8\left(\cos 8x - \tan^2 8x + 1\right)}{(1 + \sec 8x)^2}$$

##### Example 4

Suppose $$f(x) = \cot^5 11x$$. Find $$f'\left(\frac \pi 3\right)$$.

Step 1

Rewrite the function to emphasize that the cotangent is being raised to the fifth power.

$$f(x) = \left(\cot 11x\right)^5$$

Step 2

Differentiate using the chain rule.

\begin{align*} f'(x) & = 5\left(\cot 11x\right)^4\cdot \frac d {dx}\left(\cot 11x\right)\\[6pt] & = 5\left(\cot 11x\right)^4\cdot (-11\csc^2 11x)\\[6pt] & = -55\left(\cot 11x\right)^4\cdot \csc^2 11x \end{align*}

Step 3

Evaluate $$f'\left(\frac \pi 3\right)$$

\begin{align*} f'\left(\frac \pi 3\right) & = -55\left[\cot \left(11\cdot \frac \pi 3\right)\right]^4\cdot \csc^2\left(11\cdot \frac \pi 3\right)\\[6pt] & = -55\left[-\frac{\sqrt 3} 3\right]^4\cdot \left(-\frac{2\sqrt 3} 3\right)^2\\[6pt] & = -55\left[\frac 9{81}\right]\cdot \left(\frac{12} 9\right)\\[6pt] & = -\frac{220}{27} \end{align*}

$$\displaystyle f'\left(\frac\pi 3\right) = -\frac{220}{27}$$.