# Derivatives of Trigonometric Functions

The basic trigonometric functions include the following $$6$$ functions: sine $$\left(\sin x\right),$$ cosine $$\left(\cos x\right),$$ tangent $$\left(\tan x\right),$$ cotangent $$\left(\cot x\right),$$ secant $$\left(\sec x\right)$$ and cosecant $$\left(\csc x\right).$$

All these functions are continuous and differentiable in their domains. Below we make a list of derivatives for these functions.

## Derivatives of Basic Trigonometric Functions

We have already derived the derivatives of sine and cosine on the Definition of the Derivative page. They are as follows:

${\left( {\sin x} \right)^\prime } = \cos x,\;\;{\left( {\cos x} \right)^\prime } = - \sin x.$

Using the quotient rule it is easy to obtain an expression for the derivative of tangent:

$\left( {\tan x} \right)^\prime = \left( {\frac{{\sin x}}{{\cos x}}} \right)^\prime = \frac{{{{\left( {\sin x} \right)}^\prime }\cos x - \sin x{{\left( {\cos x} \right)}^\prime }}}{{{{\cos }^2}x}} = \frac{{\cos x \cdot \cos x - \sin x \cdot \left( { - \sin x} \right)}}{{{{\cos }^2}x}} = \frac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\cos }^2}x}} = \frac{1}{{{{\cos }^2}x}}.$

The derivative of cotangent can be found in the same way. However, this can be also done using the chain rule for differentiating a composite function:

$\require{cancel} \left( {\cot x} \right)^\prime = \left( {\frac{1}{{\tan x}}} \right)^\prime = - \frac{1}{{{{\tan }^2}x}} \cdot {\left( {\tan x} \right)^\prime } = - \frac{1}{{\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}} \cdot \frac{1}{{{{\cos }^2}x}} = - \frac{\cancel{{{\cos }^2}x}}{{{{\sin }^2}x \cdot \cancel{{{\cos }^2}x}}} = - \frac{1}{{{{\sin }^2}x}}.$

Similarly, we find the derivatives of secant and cosecant:

$\left( {\sec x} \right)^\prime = \left( {\frac{1}{{\cos x}}} \right)^\prime = - \frac{1}{{{{\cos }^2}x}} \cdot {\left( {\cos x} \right)^\prime } = \frac{{\sin x}}{{{{\cos }^2}x}} = \frac{{\sin x}}{{\cos x}} \cdot \frac{1}{{\cos x}} = \tan x\sec x,$
$\left( {\csc x} \right)^\prime = \left( {\frac{1}{{\sin x}}} \right)^\prime = - \frac{1}{{{{\sin }^2}x}} \cdot {\left( {\sin x} \right)^\prime } = -\frac{{\cos x}}{{{{\sin }^2}x}} = -\frac{{\cos x}}{{\sin x}} \cdot \frac{1}{{\sin x}} = -\cot x\csc x.$

## Table of Derivatives of Trigonometric Functions

The table below summarizes the derivatives of $$6$$ basic trigonometric functions:

In the examples below, find the derivative of the given function.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

$y = \cos 2x - 2\sin x$

### Example 2

$y = \tan x + \frac{1}{3}{\tan ^3}x$

### Example 3

$y = \cos x - {\frac{1}{3}}{\cos ^3}x$

### Example 4

$y = \frac{1}{{{{\cos }^n}x}}$

### Example 5

$y = {\frac{{\sin x}}{{1 + \cos x}}}$

### Example 6

$y = {\cos ^2}\sin x$

### Example 1.

$y = \cos 2x - 2\sin x$

Solution.

Using the linear properties of the derivative, the chain rule and the double angle formula, we obtain:

$y'\left( x \right) = \left( {\cos 2x - 2\sin x} \right)^\prime = \left( {\cos 2x} \right)^\prime - \left( {2\sin x} \right)^\prime = \left( { - \sin 2x} \right) \cdot {\left( {2x} \right)^\prime } - 2{\left( {\sin x} \right)^\prime } = - 2\sin 2x - 2\cos x = - 4\sin x\cos x - 2\cos x = - 2\cos x\left( {2\sin x + 1} \right).$

### Example 2.

$y = \tan x + \frac{1}{3}{\tan ^3}x$

Solution.

The derivative of this function is

$y'\left( x \right) = \left( {\tan x + \frac{1}{3}{{\tan }^3}x} \right)^\prime = \left( {\tan x} \right)^\prime + \left( {\frac{1}{3}{{\tan }^3}x} \right)^\prime = \frac{1}{{{{\cos }^2}x}} + \frac{1}{3} \cdot 3{\tan ^2}x \cdot {\left( {\tan x} \right)^\prime } = \frac{1}{{{{\cos }^2}x}} + {\tan ^2}x \cdot \frac{1}{{{{\cos }^2}x}} = \frac{{1 + {{\tan }^2}x}}{{{{\cos }^2}x}}.$

The numerator can be simplified using the trigonometric identity

$1 + {\tan^2}x = {\sec ^2}x = \frac{1}{{{{\cos }^2}x}}.$

Therefore

$y'\left( x \right) = \frac{{1 + {{\tan }^2}x}}{{{{\cos }^2}x}} = \frac{{\frac{1}{{{{\cos }^2}x}}}}{{{{\cos }^2}x}} = \frac{1}{{{{\cos }^4}x}} = {\sec ^4}x.$

### Example 3.

$y = \cos x - {\frac{1}{3}}{\cos ^3}x$

Solution.

Using the power rule and the chain rule, we get

$y^\prime = \left( {\cos x - \frac{1}{3}{{\cos }^3}x} \right)^\prime = \left( {\cos x} \right)^\prime - \left( {\frac{1}{3}{{\cos }^3}x} \right)^\prime = - \sin x - \frac{1}{3} \cdot 3{\cos ^2}x \cdot \left( {\cos x} \right)^\prime = - \sin x - {\cos ^2}x \cdot \left( { - \sin x} \right) = - \sin x + {\cos ^2}x\sin x = - \sin x\left( {1 - {{\cos }^2}x} \right) = - \sin x\,{\sin ^2}x = - {\sin ^3}x.$

### Example 4.

$y = \frac{1}{{{{\cos }^n}x}}$

Solution.

We find the derivative of this function using the power rule and the chain rule:

$y'\left( x \right) = \left( {\frac{1}{{{{\cos }^n}x}}} \right)^\prime = \left[ {{{\left( {\cos x} \right)}^{ - n}}} \right]^\prime = - n{\left( {\cos x} \right)^{ - n - 1}} \cdot {\left( {\cos x} \right)^\prime } = - n{\left( {\cos x} \right)^{ - n - 1}} \cdot \left( { -\sin x} \right) = \frac{{n\sin x}}{{{{\cos }^{n + 1}}x}}.$

Here we assume that $$\cos x \ne 0$$, that is $$x \ne {\frac{\pi }{2}} + \pi n,$$ $$n \in \mathbb{Z}.$$

### Example 5.

$y = {\frac{{\sin x}}{{1 + \cos x}}}$

Solution.

By the quotient rule,

$y^\prime = \left( {\frac{{\sin x}}{{1 + \cos x}}} \right)^\prime = \frac{{\cos x \left( {1 + \cos x} \right) - \sin x \cdot \left( { - \sin x} \right)}}{{{{\left( {1 + \cos x} \right)}^2}}} = \frac{{\cos x + {{\cos }^2}x + {{\sin }^2}x}}{{{{\left( {1 + \cos x} \right)}^2}}} = \frac{\cancel{1 + \cos x}}{{{{\left( {1 + \cos x} \right)}^\cancel{2}}}} = \frac{1}{{1 + \cos x}}.$

### Example 6.

$y = {\cos ^2}\sin x$

Solution.

Applying the power rule and the chain rule, we obtain:

$y'\left( x \right) = \left( {{{\cos }^2}\sin x} \right)^\prime = 2\cos \sin x \cdot {\left( {\cos \sin x} \right)^\prime } = 2\cos \sin x \cdot \left( { - \sin\sin x} \right) \cdot {\left( {\sin x} \right)^\prime } = - 2\cos \sin x \cdot \sin \sin x \cdot \cos x.$

The last expression can be simplified by the double angle formula:

$2\cos \sin x \cdot \sin \sin x = \sin \left( {2\sin x} \right).$

Consequently, the derivative is

$y'\left( x \right) = - \sin \left( {2\sin x} \right)\cos x.$

See more problems on Page 2.