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# Calculus

Differentiation of Functions

# Derivatives of Trigonometric Functions

Page 1
Problems 1-3
Page 2
Problems 4-17

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 $$\left(\csc x\right).$$

For each of these functions, there is an inverse trigonometric function. They are called the inverse sine $$\left(\arcsin x\right),$$ inverse cosine $$\left(\arccos x\right),$$ inverse tangent $$\left(\arctan x\right),$$ inverse cotangent $$\left({\text{arccot }x}\right),$$ inverse secant $$\left({\text{arcsec }x}\right)$$ and inverse cosecant $$\left({\text{arccsc }x}\right),$$ respectively.

All these functions are continuous and differentiable in their domains. Below we make a list of derivatives for these $$12$$ 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:
${{\left( {\sin x} \right)^\prime } = \cos x,\;\;}\kern-0.3pt{{\left( {\cos x} \right)^\prime } = – \sin x.}$ 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:
${{\left( {\sec x} \right)^\prime } = {\left( {\frac{1}{{\cos x}}} \right)^\prime } } = { – \frac{1}{{{{\cos }^2}x}} \cdot {\left( {\cos x} \right)^\prime } } = {\frac{{\sin x}}{{{{\cos }^2}x}} } = {\frac{{\sin x}}{{\cos x}} \cdot \frac{1}{{\cos x}} } = {\tan x\sec x,}$ ${{\left( {\csc x} \right)^\prime } = {\left( {\frac{1}{{\sin x}}} \right)^\prime } } = { – \frac{1}{{{{\sin }^2}x}} \cdot {\left( {\sin x} \right)^\prime } } = {-\frac{{\cos x}}{{{{\sin }^2}x}} } = {-\frac{{\cos x}}{{\sin x}} \cdot \frac{1}{{\sin x}} } = {-\cot x\csc x.}$

### Derivatives of Inverse Trigonometric Functions

The derivatives of the inverse trigonometric functions can be derived using the inverse function theorem. For example, the sine function $$x = \varphi \left( y \right)$$ $$= \sin y$$ is the inverse function for $$y = f\left( x \right)$$ $$= \arcsin x.$$ Then the derivative of $$y = \arcsin x$$ is given by
${{\left( {\arcsin x} \right)^\prime } = f’\left( x \right) = \frac{1}{{\varphi’\left( y \right)}} } = {\frac{1}{{{{\left( {\sin y} \right)}^\prime }}} } = {\frac{1}{{\cos y}} } = {\frac{1}{{\sqrt {1 – {\sin^2}y} }} } = {\frac{1}{{\sqrt {1 – {\sin^2}\left( {\arcsin x} \right)} }} } = {\frac{1}{{\sqrt {1 – {x^2}} }}\;\;}\kern-0.3pt{\left( { – 1 < x < 1} \right).}$ Using this technique, we can find the derivatives of the other inverse trigonometric functions:
${{\left( {\arccos x} \right)^\prime } }={ \frac{1}{{{{\left( {\cos y} \right)}^\prime }}} } = {\frac{1}{{\left( { – \sin y} \right)}} } = {- \frac{1}{{\sqrt {1 – {{\cos }^2}y} }} } = {- \frac{1}{{\sqrt {1 – {{\cos }^2}\left( {\arccos x} \right)} }} } = {- \frac{1}{{\sqrt {1 – {x^2}} }}\;\;}\kern-0.3pt{\left( { – 1 < x < 1} \right),}\qquad$ ${{\left( {\arctan x} \right)^\prime } }={ \frac{1}{{{{\left( {\tan y} \right)}^\prime }}} } = {\frac{1}{{\frac{1}{{{{\cos }^2}y}}}} } = {\frac{1}{{1 + {{\tan }^2}y}} } = {\frac{1}{{1 + {{\tan }^2}\left( {\arctan x} \right)}} } = {\frac{1}{{1 + {x^2}}},}$ ${\left( {\text{arccot }x} \right)^\prime } = {\frac{1}{{{{\left( {\cot y} \right)}^\prime }}}} = \frac{1}{{\left( { – \frac{1}{{{\sin^2}y}}} \right)}} = – \frac{1}{{1 + {{\cot }^2}y}} = – \frac{1}{{1 + {{\cot }^2}\left( {\text{arccot }x} \right)}} = – \frac{1}{{1 + {x^2}}},$ ${{\left( {\text{arcsec }x} \right)^\prime } = {\frac{1}{{{{\left( {\sec y} \right)}^\prime }}} }} = {\frac{1}{{\tan y\sec y}} } = {\frac{1}{{\sec y\sqrt {{{\sec }^2}y – 1} }} } = {\frac{1}{{\left| x \right|\sqrt {{x^2} – 1} }}.}$ In the last formula, the absolute value $$\left| x \right|$$ in the denominator appears due to the fact that the product $${\tan y\sec y}$$ should always be positive in the range of admissible values of $$y$$, where $$y \in \left( {0,{\large\frac{\pi }{2}\normalsize}} \right) \cup \left( {{\large\frac{\pi }{2}\normalsize},\pi } \right),$$ i.e. the derivative of the inverse secant is always positive.

Similarly, we can obtain an expression for the derivative of the inverse cosecant function:
${{\left( {\text{arccsc }x} \right)^\prime } = {\frac{1}{{{{\left( {\csc y} \right)}^\prime }}} }} = {-\frac{1}{{\cot y\csc y}} } = {-\frac{1}{{\csc y\sqrt {{{\csc }^2}y – 1} }} } = {-\frac{1}{{\left| x \right|\sqrt {{x^2} – 1} }}.}$

### Table of Derivatives of Trigonometric Functions

The derivatives of $$6$$ basic trigonometric functions and $$6$$ inverse trigonometric functions considered above are presented in the following table:

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

## Solved Problems

Click on problem description to see solution.

### ✓Example 1

$y = \cos 2x – 2\sin x$

### ✓Example 2

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

### ✓Example 3

${y = \arctan \frac{{x + 1}}{{x – 1}}\;\;}\kern-0.3pt{\left( {x \ne 1} \right)}$

### ✓Example 4

$y = \arctan \left( {x – \sqrt {1 + {x^2}} } \right)$

### ✓Example 5

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

### ✓Example 6

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

### ✓Example 7

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

### ✓Example 8

$y = {\sin ^3}x + {\cos ^3}x$

### ✓Example 9

$y = \tan \frac{x}{2} – \cot \frac{x}{2}$

### ✓Example 10

Using the chain rule, derive the formula for the derivative of the inverse sine function.

### ✓Example 11

$y = \arcsin \sqrt {1 – {x^2}}$

### ✓Example 12

$y = \frac{2}{{\sqrt 3 }}\text{arccot}\frac{{2x + 1}}{{\sqrt 3 }}$

### ✓Example 13

$y = {\sin ^n}x\cos nx$

### ✓Example 14

Show that $${\large\frac{d}{{dx}}\normalsize} \left( {x\arcsin x + \sqrt {1 – {x^2}} } \right)$$ $$= \arcsin x.$$

### ✓Example 15

${y = {\left( {\tan x} \right)^{\cos x}},\;\;\;}\kern0pt {\text{where}\;\;0 \lt x \lt \frac{\pi }{2}.}$

### ✓Example 16

$y = \arcsin \frac{{1 – {x^2}}}{{1 + {x^2}}}$

### ✓Example 17

$y = \frac{1}{x}\text{arccsc}\frac{1}{x}$

### 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:
${y’\left( x \right) }={ {\left( {\cos 2x – 2\sin x} \right)^\prime } } = {{\left( {\cos 2x} \right)^\prime } – {\left( {2\sin x} \right)^\prime } } = {\left( { – \sin 2x} \right) \cdot {\left( {2x} \right)^\prime } – 2{\left( {\sin x} \right)^\prime } } = { – 2\sin 2x – 2\cos x } = { – 2\sin x\cos x – 2\cos x } = { – 2\cos x\left( {\sin x + 1} \right).}$

### Example 2.

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

#### Solution.

The derivative of this function is as follows:
${y’\left( x \right) }={ {\left( {\tan x + \frac{1}{3}{{\tan }^3}x} \right)^\prime } } = {{\left( {\tan x} \right)^\prime } + {\left( {\frac{1}{3}{{\tan }^3}x} \right)^\prime } } = {\frac{1}{{{{\cos }^2}x}} + \frac{1}{3} \cdot 3{\tan ^2}x \cdot {\left( {\tan x} \right)^\prime } } = {\frac{1}{{{{\cos }^2}x}} + {\tan ^2}x \cdot \frac{1}{{{{\cos }^2}x}} } = {\frac{{1 + {{\tan }^2}x}}{{{{\cos }^2}x}}.}$ 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.

${y = \arctan \frac{{x + 1}}{{x – 1}}\;\;}\kern-0.3pt{\left( {x \ne 1} \right)}$

#### Solution.

By the chain and quotient rules, we have
${y’\left( x \right) }={ {\left( {\arctan \frac{{x + 1}}{{x – 1}}} \right)^\prime } } = {\frac{1}{{1 + {{\left( {\frac{{x + 1}}{{x – 1}}} \right)}^2}}} \cdot {\left( {\frac{{x + 1}}{{x – 1}}} \right)^\prime } } = {\frac{{1 \cdot \left( {x – 1} \right) – \left( {x + 1} \right) \cdot 1}}{{{{\left( {x – 1} \right)}^2} + {{\left( {x + 1} \right)}^2}}} } = {\frac{{\cancel{\color{blue}{x}} – \color{red}{1} – \cancel{\color{blue}{x}} – \color{red}{1}}}{{\color{maroon}{x^2} – \cancel{\color{green}{2x}} + \color{DarkViolet}{1} + \color{maroon}{x^2} + \cancel{\color{green}{2x}} + \color{DarkViolet}{1}}} } = {\frac{{ – \color{red}{2}}}{{\color{maroon}{2{x^2}} + \color{DarkViolet}{2}}} } = { – \frac{1}{{1 + {x^2}}}.}$

Page 1
Problems 1-3
Page 2
Problems 4-17