The Chain Rule, Differentiating Reciprocal Trigonometric Functions — Edexcel A-level Mathematics (Pure)

SPEC 7.4 • DIFFERENTIATION

The Chain Rule & Reciprocal Trig Functions

Differentiating composite functions and sec x, cosec x, cot x

Composite Functions

A composite function is a "function inside a function" — written as

f (g (x))

.

Examples

sin (x^{2})

— sine of

x^{2}

(trig of polynomial)

e^{3 x + 1}

— exponential of

3 x + 1

(exp of linear)

(2 x + 5)^{7}

— power of a linear function

ln (cos x)

— ln of cosine (log of trig)

Key Idea

We can't differentiate these using basic rules alone. We need the Chain Rule to handle the "inner" function.

The Chain Rule

\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}

If

y = f (u)

and

u = g (x)

, then differentiate

y

with respect to

u

, then multiply by the derivative of

u

.

Memory Trick

"Differentiate the OUTER function, then multiply by the derivative of the INNER function."

Alternative Notation

If

y = f (g (x))

, then:

\frac{d y}{d x} = f^{'} (g (x)) \times g^{'} (x)

Chain Rule with Powers

For functions of the form

y = (f (x))^{n}

:

Formula

\frac{d}{d x} [f (x)]^{n} = n [f (x)]^{n - 1} \times f^{'} (x)

Example 1

y = (3 x - 2)^{5}

\frac{d y}{d x} = 5 (3 x - 2)^{4} \times 3

= 15 (3 x - 2)^{4}

Example 2

y = 2 x + 7 = (2 x + 7)^{1/2}

\frac{d y}{d x} = \frac{1}{2} (2 x + 7)^{- 1/2} \times 2

= \frac{1}{2 x + 7}

Worked Example01

Differentiate $y = (4 x - 3)^{6}$

1

Identify: outer function is

u^{6}

, inner function is

u = 4 x - 3

2

Differentiate outer:

\frac{d}{d u} (u^{6}) = 6 u^{5} = 6 (4 x - 3)^{5}

3

Differentiate inner:

\frac{d}{d x} (4 x - 3) = 4

4

Multiply:

\frac{d y}{d x} = 6 (4 x - 3)^{5} \times 4 = 24 (4 x - 3)^{5}

Answer: $\frac{d y}{d x} = 24 (4 x - 3)^{5}$

Worked Example02

Differentiate $y = 1 - x^{2}$

1

Rewrite:

y = (1 - x^{2})^{1/2}

2

Outer: power

\frac{1}{2}

, inner:

1 - x^{2}

3

\frac{d y}{d x} = \frac{1}{2} (1 - x^{2})^{- 1/2} \times (- 2 x)

4

Simplify:

\frac{d y}{d x} = \frac{- x}{1 - x ^{2}}

Answer: $\frac{d y}{d x} = \frac{- x}{1 - x ^{2}}$

Worked Example03

Differentiate $y = \frac{1}{s i n ( 2 x )}$

1

Rewrite:

y = [sin (2 x)]^{- 1/2}

2

Outer: power

- \frac{1}{2}

, inner:

sin (2 x)

3

Differentiate outer:

- \frac{1}{2} [sin (2 x)]^{- 3/2}

4

Differentiate inner:

\frac{d}{d x} [sin (2 x)] = 2 cos (2 x)

5

Multiply:

\frac{d y}{d x} = - \frac{1}{2} [sin (2 x)]^{- 3/2} \times 2 cos (2 x)

6

Simplify:

= \frac{- c o s ( 2 x )}{[ s i n ( 2 x ) ] ^{3/2}}

Answer: $\frac{d y}{d x} = \frac{- c o s ( 2 x )}{[ s i n ( 2 x ) ] ^{3/2}}$

Chain Rule with Exponentials

For functions of the form

y = e^{f (x)}

:

Formula

\frac{d}{d x} [e^{f (x)}] = f^{'} (x) \times e^{f (x)}

Example 1

y = e^{3 x}

\frac{d y}{d x} = 3 e^{3 x}

Example 2

y = e^{x^{2} + 1}

\frac{d y}{d x} = 2 x \cdot e^{x^{2} + 1}

Remember

The exponential

e^{f (x)}

stays exactly the same — you just multiply by the derivative of the power!

Worked Example04

Differentiate $y = e^{x^{2} - 2 x}$

1

Identify: outer is

e^{u}

, inner is

u = x^{2} - 2 x

2

Differentiate inner:

\frac{d u}{d x} = 2 x - 2

3

Apply chain rule:

\frac{d y}{d x} = (2 x - 2) \times e^{x^{2} - 2 x}

4

Simplify:

\frac{d y}{d x} = 2 (x - 1) e^{x^{2} - 2 x}

Answer: $\frac{d y}{d x} = 2 (x - 1) e^{x^{2} - 2 x}$

Chain Rule with Trig Functions

For functions like

sin (f (x))

,

cos (f (x))

,

tan (f (x))

:

Formulas

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

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

\frac{d}{d x} [tan (f (x))] = f^{'} (x) sec^{2} (f (x))

Example

y = sin (3 x)

\frac{d y}{d x} = 3 cos (3 x)

Example

y = cos (x^{2})

\frac{d y}{d x} = - 2 x sin (x^{2})

Worked Example05

Differentiate $y = sin (5 x + 2)$

1

Identify: outer is

sin (u)

, inner is

u = 5 x + 2

2

Differentiate outer:

\frac{d}{d u} (sin u) = cos u = cos (5 x + 2)

3

Differentiate inner:

\frac{d}{d x} (5 x + 2) = 5

4

Multiply:

\frac{d y}{d x} = 5 cos (5 x + 2)

Answer: $\frac{d y}{d x} = 5 cos (5 x + 2)$

Chain Rule with Logarithms

For functions of the form

y = ln (f (x))

:

Formula

\frac{d}{d x} [ln (f (x))] = \frac{f ^{'} ( x )}{f ( x )}

Example 1

y = ln (3 x - 2)

\frac{d y}{d x} = \frac{3}{3 x - 2}

Example 2

y = ln (x^{2} + 1)

\frac{d y}{d x} = \frac{2 x}{x ^{2} + 1}

Pattern

Derivative of

ln (something)

=

\frac{derivative of something}{something}

Worked Example06

Differentiate $y = ln (x^{3} + 2 x)$

1

Identify:

f (x) = x^{3} + 2 x

2

Find

f^{'} (x) = 3 x^{2} + 2

3

Apply formula:

\frac{d y}{d x} = \frac{f ^{'} ( x )}{f ( x )} = \frac{3 x ^{2} + 2}{x ^{3} + 2 x}

Answer: $\frac{d y}{d x} = \frac{3 x ^{2} + 2}{x ^{3} + 2 x}$

Reciprocal Trig Functions

These are the reciprocals of sine, cosine, and tangent:

Definitions

sec x = \frac{1}{c o s x}

(reciprocal of cosine)

csc x = \frac{1}{s i n x}

(reciprocal of sine)

cot x = \frac{1}{t a n x} = \frac{c o s x}{s i n x}

(reciprocal of tangent)

Memory Aid

The third letter tells you the reciprocal: sec → 1/cos, cosec → 1/sin, cot → 1/tan

Deriving: Derivative of sec x

We use the chain rule to derive the derivative of

sec x

:

Derivation

Step 1: Write

sec x = (cos x)^{- 1}

Step 2: Apply chain rule with outer

u^{- 1}

and inner

u = cos x

Step 3:

\frac{d}{d x} (sec x) = - 1 (cos x)^{- 2} \times (- sin x)

Step 4:

= \frac{s i n x}{c o s ^{2} x} = \frac{1}{c o s x} \times \frac{s i n x}{c o s x}

Result

\frac{d}{d x} (sec x) = sec x tan x

Deriving: Derivative of csc x

We use the chain rule to derive the derivative of

csc x

:

Derivation

Step 1: Write

csc x = (sin x)^{- 1}

Step 2: Apply chain rule with outer

u^{- 1}

and inner

u = sin x

Step 3:

\frac{d}{d x} (csc x) = - 1 (sin x)^{- 2} \times (cos x)

Step 4:

= \frac{- c o s x}{s i n ^{2} x} = - \frac{1}{s i n x} \times \frac{c o s x}{s i n x}

Result

\frac{d}{d x} (csc x) = - csc x cot x

Deriving: Derivative of cot x

We use the quotient rule or chain rule to derive the derivative of

cot x

:

Derivation (Using Chain Rule)

Step 1: Write

cot x = (tan x)^{- 1}

Step 2: Apply chain rule with outer

u^{- 1}

and inner

u = tan x

Step 3:

\frac{d}{d x} (cot x) = - 1 (tan x)^{- 2} \times sec^{2} x

Step 4:

= \frac{- s e c ^{2} x}{t a n ^{2} x} = \frac{- 1/ c o s ^{2} x}{s i n ^{2} x / c o s ^{2} x} = \frac{- 1}{s i n ^{2} x}

Result

\frac{d}{d x} (cot x) = - csc^{2} x

Reciprocal Trig Derivatives

Key Results

\frac{d}{d x} (sec x) = sec x tan x

\frac{d}{d x} (csc x) = - csc x cot x

\frac{d}{d x} (cot x) = - csc^{2} x

Watch the Signs

Both cosec and cot have negative derivatives! Only sec x has a positive derivative.

With Chain Rule

For

sec (f (x))

: derivative is

f^{'} (x) sec (f (x)) tan (f (x))

For

csc (f (x))

: derivative is

- f^{'} (x) csc (f (x)) cot (f (x))

For

cot (f (x))

: derivative is

- f^{'} (x) csc^{2} (f (x))

Worked Example07

Differentiate $y = sec (3 x)$

1

Use:

\frac{d}{d x} (sec u) = sec u tan u \times \frac{d u}{d x}

2

Here

u = 3 x

, so

\frac{d u}{d x} = 3

3

\frac{d y}{d x} = sec (3 x) tan (3 x) \times 3

4

= 3 sec (3 x) tan (3 x)

Answer: $\frac{d y}{d x} = 3 sec (3 x) tan (3 x)$

Worked Example08

Differentiate $y = csc (2 x + 1)$

1

Use:

\frac{d}{d x} (csc u) = - csc u cot u \times \frac{d u}{d x}

2

Here

u = 2 x + 1

, so

\frac{d u}{d x} = 2

3

\frac{d y}{d x} = - csc (2 x + 1) cot (2 x + 1) \times 2

4

= - 2 csc (2 x + 1) cot (2 x + 1)

Answer: $\frac{d y}{d x} = - 2 csc (2 x + 1) cot (2 x + 1)$

Worked Example09

Differentiate $y = [cot (2 x^{2} - 1)]^{3/2}$

1

This requires TWO applications of the chain rule (nested functions)

2

Outermost: power

\frac{3}{2}

, middle:

cot (u)

, innermost:

u = 2 x^{2} - 1

3

Differentiate outer:

\frac{3}{2} [cot (2 x^{2} - 1)]^{1/2}

4

Multiply by derivative of

cot (2 x^{2} - 1)

:

= - csc^{2} (2 x^{2} - 1) \times 4 x

5

Combine:

\frac{d y}{d x} = \frac{3}{2} [cot (2 x^{2} - 1)]^{1/2} \times (- 4 x) csc^{2} (2 x^{2} - 1)

6

Simplify:

= - 6 x csc^{2} (2 x^{2} - 1) cot (2 x^{2} - 1)

Answer: $\frac{d y}{d x} = - 6 x csc^{2} (2 x^{2} - 1) cot (2 x^{2} - 1)$

Practice

Question 1: Differentiate

y = (2 x + 7)^{8}

Using chain rule with

u = 2 x + 7

:

\frac{d y}{d x} = 8 (2 x + 7)^{7} \times 2 = 16 (2 x + 7)^{7}

Answer: $\frac{d y}{d x} = 16 (2 x + 7)^{7}$

Practice

Question 2: Differentiate

y = cos^{3} (x)

[Hint: This means

(cos x)^{3}

]

Rewrite as

y = (cos x)^{3}

. Using chain rule:

Outer: power 3, inner:

cos x

\frac{d y}{d x} = 3 (cos x)^{2} \times (- sin x)

= - 3 cos^{2} x sin x

Answer: $\frac{d y}{d x} = - 3 cos^{2} x sin x$

Practice

Question 3: Differentiate

y = e^{t a n x}

Using chain rule for

e^{f (x)}

:

Inner function:

f (x) = tan x

, so

f^{'} (x) = sec^{2} x

\frac{d y}{d x} = sec^{2} x \times e^{t a n x}

Answer: $\frac{d y}{d x} = sec^{2} x \cdot e^{t a n x}$

Summary

Chain Rule

Function TypeDerivative

[f (x)]^{n}

n [f (x)]^{n - 1} \times f^{'} (x)

e^{f (x)}

f^{'} (x) \times e^{f (x)}

sin (f (x))

f^{'} (x) cos (f (x))

cos (f (x))

- f^{'} (x) sin (f (x))

ln (f (x))

\frac{f ^{'} ( x )}{f ( x )}

Reciprocal Trig Derivatives

\frac{d}{d x} (sec x) = sec x tan x

|

\frac{d}{d x} (csc x) = - csc x cot x

|

\frac{d}{d x} (cot x) = - csc^{2} x