#Study Notes: Derivatives of Trigonometric Functions

#Table of Contents

Derivative of the Tangent Function
Derivatives of Reciprocal Trigonometric Functions
Practice Questions
Glossary
Summary and Key Takeaways
Exam Strategy

#Derivative of the Tangent Function

#What is the Derivative of $\tan x$ ?

Key Concept

If $f(x) = \tan x$ , then the derivative is given by: $f'(x) = \sec^2 x$

This can be shown using the identity $\tan x \equiv \frac{\sin x}{\cos x}$ and the quotient rule.

The quotient rule states that if $y = \frac{u}{v}$ , then $y' = \frac{u'v - uv'}{v^2}$

Let

u = \sin x

and

v = \cos x

. -

u' = \cos x

v' = -\sin x

Applying the quotient rule: $y' = \frac{\cos x \cdot \cos x + \sin x \cdot \sin x}{\cos^2 x}$ Simplifying: $y' = \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$ Using the identity $\sin^2 x + \cos^2 x \equiv 1$ : $y' = \frac{1}{\cos^2 x} = \sec^2 x$

#What is the Derivative of $\tan kx$ ?

Key Concept

If $f(x) = \tan kx$ , then the derivative is: $f'(x) = k \sec^2 kx$

This is a result of applying the chain rule. It can also be shown using the quotient rule for $\frac{\sin kx}{\cos kx}$ .

#Worked Examples

Differentiate the following functions:

(a) $f(x) = \tan 3x$

Answer:

Using the rule for $\tan kx$ : $f'(x) = 3 \sec^2 3x$

(b) $g(x) = 3x^2 \tan x$

Answer:

This is a product of two terms, so use the product rule: $y' = u'v + uv'$

Let $u = 3x^2$ and $v = \tan x$ .

$u' = 6x$
$v' = \sec^2 x$

Applying the product rule: $y' = 6x \cdot \tan x + 3x^2 \cdot \sec^2 x = 3x (2 \tan x + x \sec^2 x)$

#Derivatives of Reciprocal Trigonometric Functions

#What are the Reciprocal Trig Functions?

Key Concept

The reciprocal trigonometric functions are:

$\csc x = \frac{1}{\sin x}$
$\sec x = \frac{1}{\cos x}$
$\cot x = \frac{1}{\tan x}$

**csc** is sometimes written as **cosec**. You can remember the functions by looking at the **third letter** of each: - co**s**ec is the reciprocal of **s**in - se**c** is the reciprocal of **c**os - co**t** is the reciprocal of **t**an

#What are the Derivatives of the Reciprocal Trig Functions?

Key Concept

If $f(x) = \csc x$ , then: $f'(x) = -\cot x \csc x$

If $g(x) = \sec x$ , then: $g'(x) = \tan x \sec x$

If $h(x) = \cot x$ , then: $h'(x) = -\csc^2 x$

These results can be remembered or derived using the reciprocal trig function definitions and the quotient rule.

#How to Derive the Derivatives

#Derivative of $\csc x$

Recall that $\csc x = \frac{1}{\sin x}$ . Apply the quotient rule: $y' = \frac{u'v - uv'}{v^2}$

Let

u = 1

and

v = \sin x

. -

u' = 0

v' = \cos x

Applying the quotient rule: $y' = \frac{0 - \cos x}{(\sin x)^2} = \frac{-\cos x}{\sin^2 x}$

Simplify using the identities $\csc x = \frac{1}{\sin x}$ and $\cot x = \frac{\cos x}{\sin x}$ : $y' = -\cot x \csc x$

#Derivative of $\sec x$

Recall that $\sec x = \frac{1}{\cos x}$ . Apply the quotient rule: $y' = \frac{u'v - uv'}{v^2}$

Let

u = 1

and

v = \cos x

. -

u' = 0

v' = -\sin x

Applying the quotient rule: $y' = \frac{0 - (-\sin x)}{(\cos x)^2} = \frac{\sin x}{\cos^2 x}$

Simplify using the identity $\sec x = \frac{1}{\cos x}$ : $y' = \tan x \sec x$

#Derivative of $\cot x$

Recall that $\cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x}$ . Apply the quotient rule: $y' = \frac{u'v - uv'}{v^2}$

Let

u = \cos x

and

v = \sin x

. -

u' = -\sin x

v' = \cos x

Applying the quotient rule: $y' = \frac{-\sin x \cdot \sin x - \cos x \cdot \cos x}{(\sin x)^2} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x}$

Simplify using the identity $\sin^2 x + \cos^2 x = 1$ : $y' = \frac{-1}{\sin^2 x} = -\csc^2 x$

#Worked Example

Show that the derivative of

f(x) = \csc x \sec x

f'(x) = \sec^2 x - \csc^2 x

This is a product of two functions, so use the product rule: $y' = u'v + uv'$

Let $u = \csc x$ and $v = \sec x$ . Differentiate using the known results:

$u' = -\cot x \csc x$
$v' = \tan x \sec x$

Apply the product rule: $y' = -\cot x \csc x \cdot \sec x + \csc x \cdot \tan x \sec x$

Using trigonometric identities to simplify: $y' = \left( \frac{-\cos x}{\sin x} \cdot \frac{1}{\sin x} \cdot \frac{1}{\cos x} \right) + \left( \frac{1}{\sin x} \cdot \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} \right)$

Simplify further: $y' = \frac{-1}{\sin^2 x} + \frac{1}{\cos^2 x} = -\csc^2 x + \sec^2 x$

#Practice Questions

Practice Question

Differentiate $f(x) = \tan 5x$ .
Differentiate $g(x) = x^2 \sec x$ .
Show that the derivative of $h(x) = \cot 2x$ is $h'(x) = -2 \csc^2 2x$ .
Differentiate $k(x) = 2x \csc x$ .

#Glossary

Quotient Rule: A rule for differentiating the quotient of two functions.
Chain Rule: A rule for differentiating compositions of functions.
Secant ( $\sec$ ): Reciprocal of cosine, $\sec x = \frac{1}{\cos x}$ .
Cosecant ( $\csc$ ): Reciprocal of sine, $\csc x = \frac{1}{\sin x}$ .
Cotangent ( $\cot$ ): Reciprocal of tangent, $\cot x = \frac{1}{\tan x}$ .

#Summary and Key Takeaways

The derivative of $\tan x$ is $\sec^2 x$ .
The derivative of $\tan kx$ is $k \sec^2 kx$ .
The derivatives of reciprocal trig functions can be derived using the quotient rule.
Memorizing the derivatives of $\sec x$ , $\csc x$ , and $\cot x$ is useful for solving problems quickly.

#Exam Strategy

Exam Tip

Always start by identifying which differentiation rule applies: product rule, quotient rule, or chain rule.
Simplify trigonometric expressions using identities before differentiating.
Practice deriving the formulas for derivatives of reciprocal trigonometric functions to reinforce understanding.

By following these notes and practicing the provided questions, you'll be well-prepared for questions involving the derivatives of trigonometric functions on your exam.

#Study Notes: Derivatives of Trigonometric Functions

#Table of Contents

Derivative of the Tangent Function
Derivatives of Reciprocal Trigonometric Functions
Practice Questions
Glossary
Summary and Key Takeaways
Exam Strategy

#Derivative of the Tangent Function

#What is the Derivative of $\tan x$ ?

Key Concept

If $f(x) = \tan x$ , then the derivative is given by: $f'(x) = \sec^2 x$

This can be shown using the identity $\tan x \equiv \frac{\sin x}{\cos x}$ and the quotient rule.

The quotient rule states that if $y = \frac{u}{v}$ , then $y' = \frac{u'v - uv'}{v^2}$

Let

u = \sin x

and

v = \cos x

. -

u' = \cos x

v' = -\sin x

#What is the Derivative of $\tan kx$ ?

Key Concept

If $f(x) = \tan kx$ , then the derivative is: $f'(x) = k \sec^2 kx$

This is a result of applying the chain rule. It can also be shown using the quotient rule for $\frac{\sin kx}{\cos kx}$ .

#Worked Examples

Differentiate the following functions:

(a) $f(x) = \tan 3x$

Answer:

Using the rule for $\tan kx$ : $f'(x) = 3 \sec^2 3x$

(b) $g(x) = 3x^2 \tan x$

Answer:

This is a product of two terms, so use the product rule: $y' = u'v + uv'$

Let $u = 3x^2$ and $v = \tan x$ .

$u' = 6x$
$v' = \sec^2 x$

Applying the product rule: $y' = 6x \cdot \tan x + 3x^2 \cdot \sec^2 x = 3x (2 \tan x + x \sec^2 x)$

#Derivatives of Reciprocal Trigonometric Functions

#What are the Reciprocal Trig Functions?

Key Concept

The reciprocal trigonometric functions are:

$\csc x = \frac{1}{\sin x}$
$\sec x = \frac{1}{\cos x}$
$\cot x = \frac{1}{\tan x}$

#What are the Derivatives of the Reciprocal Trig Functions?

Key Concept

If $f(x) = \csc x$ , then: $f'(x) = -\cot x \csc x$

If $g(x) = \sec x$ , then: $g'(x) = \tan x \sec x$

If $h(x) = \cot x$ , then: $h'(x) = -\csc^2 x$

These results can be remembered or derived using the reciprocal trig function definitions and the quotient rule.

#How to Derive the Derivatives

#Derivative of $\csc x$

Recall that $\csc x = \frac{1}{\sin x}$ . Apply the quotient rule: $y' = \frac{u'v - uv'}{v^2}$

Let

u = 1

and

v = \sin x

. -

u' = 0

v' = \cos x

Applying the quotient rule: $y' = \frac{0 - \cos x}{(\sin x)^2} = \frac{-\cos x}{\sin^2 x}$

Simplify using the identities $\csc x = \frac{1}{\sin x}$ and $\cot x = \frac{\cos x}{\sin x}$ : $y' = -\cot x \csc x$

#Derivative of $\sec x$

Recall that $\sec x = \frac{1}{\cos x}$ . Apply the quotient rule: $y' = \frac{u'v - uv'}{v^2}$

Let

u = 1

and

v = \cos x

. -

u' = 0

v' = -\sin x

Applying the quotient rule: $y' = \frac{0 - (-\sin x)}{(\cos x)^2} = \frac{\sin x}{\cos^2 x}$

Simplify using the identity $\sec x = \frac{1}{\cos x}$ : $y' = \tan x \sec x$

#Derivative of $\cot x$

Recall that $\cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x}$ . Apply the quotient rule: $y' = \frac{u'v - uv'}{v^2}$

Let

u = \cos x

and

v = \sin x

. -

u' = -\sin x

v' = \cos x

Applying the quotient rule: $y' = \frac{-\sin x \cdot \sin x - \cos x \cdot \cos x}{(\sin x)^2} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x}$

Simplify using the identity $\sin^2 x + \cos^2 x = 1$ : $y' = \frac{-1}{\sin^2 x} = -\csc^2 x$

#Worked Example

Show that the derivative of

f(x) = \csc x \sec x

f'(x) = \sec^2 x - \csc^2 x

This is a product of two functions, so use the product rule: $y' = u'v + uv'$

Let $u = \csc x$ and $v = \sec x$ . Differentiate using the known results:

$u' = -\cot x \csc x$
$v' = \tan x \sec x$

Apply the product rule: $y' = -\cot x \csc x \cdot \sec x + \csc x \cdot \tan x \sec x$

Simplify further: $y' = \frac{-1}{\sin^2 x} + \frac{1}{\cos^2 x} = -\csc^2 x + \sec^2 x$

#Practice Questions

Practice Question

Differentiate $f(x) = \tan 5x$ .
Differentiate $g(x) = x^2 \sec x$ .
Show that the derivative of $h(x) = \cot 2x$ is $h'(x) = -2 \csc^2 2x$ .
Differentiate $k(x) = 2x \csc x$ .

#Glossary

Quotient Rule: A rule for differentiating the quotient of two functions.
Chain Rule: A rule for differentiating compositions of functions.
Secant ( $\sec$ ): Reciprocal of cosine, $\sec x = \frac{1}{\cos x}$ .
Cosecant ( $\csc$ ): Reciprocal of sine, $\csc x = \frac{1}{\sin x}$ .
Cotangent ( $\cot$ ): Reciprocal of tangent, $\cot x = \frac{1}{\tan x}$ .

#Summary and Key Takeaways

The derivative of $\tan x$ is $\sec^2 x$ .
The derivative of $\tan kx$ is $k \sec^2 kx$ .
The derivatives of reciprocal trig functions can be derived using the quotient rule.
Memorizing the derivatives of $\sec x$ , $\csc x$ , and $\cot x$ is useful for solving problems quickly.

#Exam Strategy

Exam Tip

Always start by identifying which differentiation rule applies: product rule, quotient rule, or chain rule.
Simplify trigonometric expressions using identities before differentiating.
Practice deriving the formulas for derivatives of reciprocal trigonometric functions to reinforce understanding.

By following these notes and practicing the provided questions, you'll be well-prepared for questions involving the derivatives of trigonometric functions on your exam.

Fundamental Properties of Differentiation

Michael Green

#Study Notes: Derivatives of Trigonometric Functions

#Table of Contents

#Derivative of the Tangent Function

#What is the Derivative of tan⁡x\tan xtanx?

#What is the Derivative of tan⁡kx\tan kxtankx?

#Worked Examples

#Derivatives of Reciprocal Trigonometric Functions

#What are the Reciprocal Trig Functions?

#What are the Derivatives of the Reciprocal Trig Functions?

#How to Derive the Derivatives

#Derivative of csc⁡x\csc xcscx

#Derivative of sec⁡x\sec xsecx

#Derivative of cot⁡x\cot xcotx

#Worked Example

#Practice Questions

#Glossary

#Summary and Key Takeaways

#Exam Strategy

Continue your learning journey

How are we doing?

Fundamental Properties of Differentiation

Michael Green

#Study Notes: Derivatives of Trigonometric Functions

#Table of Contents

#Derivative of the Tangent Function

#What is the Derivative of tan⁡x\tan xtanx?

#What is the Derivative of tan⁡kx\tan kxtankx?

#Worked Examples

#Derivatives of Reciprocal Trigonometric Functions

#What are the Reciprocal Trig Functions?

#What are the Derivatives of the Reciprocal Trig Functions?

#How to Derive the Derivatives

#Derivative of csc⁡x\csc xcscx

#Derivative of sec⁡x\sec xsecx

#Derivative of cot⁡x\cot xcotx

#Worked Example

#Practice Questions

#Glossary

#Summary and Key Takeaways

#Exam Strategy

Continue your learning journey

How are we doing?

#What is the Derivative of $\tan x$ ?

#What is the Derivative of $\tan kx$ ?

#Derivative of $\csc x$

#Derivative of $\sec x$

#Derivative of $\cot x$

#What is the Derivative of $\tan x$ ?

#What is the Derivative of $\tan kx$ ?

#Derivative of $\csc x$

#Derivative of $\sec x$

#Derivative of $\cot x$