Differentiating Inverse Functions

Learn to differentiate inverse functions including arcsin, arccos, and arctan

SPEC 7.5 • DIFFERENTIATION

Inverse Functions Recap

When we have a function

y = f (x)

, its inverse function

x = f^{- 1} (y)

"undoes" the original function.

Key Relationship

If

y = f (x)

, then the inverse gives us

x = f^{- 1} (y)

Applying both:

f^{- 1} (f (x)) = x

Example

If

y = x^{3}

Then

x = y^{1/3} = 3 y

The inverse is

f^{- 1} (y) = 3 y

Graphically

The graphs of

y = f (x)

and

y = f^{- 1} (x)

are reflections in the line

y = x

The Inverse Function Derivative

There is a powerful relationship between the derivative of a function and the derivative of its inverse:

Key Formula

\frac{d y}{d x} = \frac{1}{\frac{d x}{d y}}

In Words

The derivative of

y

with respect to

x

is the reciprocal of the derivative of

x

with respect to

y

.

Why This Works

Think of

\frac{d y}{d x}

and

\frac{d x}{d y}

as fractions. If you flip

\frac{d x}{d y}

upside down, you get

\frac{d y}{d x}

.

Worked Example01

Given $y = x^{3}$ , find $\frac{d y}{d x}$ using the inverse function method.

•

1. Write

x

in terms of

y

:

x = y^{1/3}

•

2. Find

\frac{d x}{d y}

:

\frac{d x}{d y} = \frac{1}{3} y^{- 2/3} = \frac{1}{3 y ^{2/3}}

•

3. Apply the formula:

\frac{d y}{d x} = \frac{1}{\frac{d x}{d y}} = \frac{1}{\frac{1}{3 y ^{2/3}}} = 3 y^{2/3}

•

4. Substitute back

y = x^{3}

:

\frac{d y}{d x} = 3 (x^{3})^{2/3} = 3 x^{2}

Answer: $\frac{d y}{d x} = 3 x^{2}$ (which matches the standard result!)

Worked Example02

If $x = sin y$ , find $\frac{d y}{d x}$ .

•

1. Note that

y = arcsin x

(the inverse of

x = sin y

)

•

2. Find

\frac{d x}{d y}

: Since

x = sin y

, we have

\frac{d x}{d y} = cos y

•

3. Apply the formula:

\frac{d y}{d x} = \frac{1}{\frac{d x}{d y}} = \frac{1}{c o s y}

•

4. Express in terms of

x

: Using

sin^{2} y + cos^{2} y = 1

, we get

cos y = 1 - sin^{2} y = 1 - x^{2}

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

Derivative of arcsin x

Let's derive the formula for differentiating

y = arcsin x

step by step:

Start with

y = arcsin x

, which means

x = sin y

Differentiate both sides with respect to

y

:

\frac{d x}{d y} = cos y

Apply the inverse formula:

\frac{d y}{d x} = \frac{1}{c o s y}

Use identity

cos y = 1 - sin^{2} y = 1 - x^{2}

Standard Result

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

Domain

This formula is only valid for

- 1 < x < 1

(the domain of arcsin)

Worked Example03

Differentiate $y = arcsin (2 x)$ .

•

1. Use the chain rule: if

y = arcsin (u)

where

u = 2 x

•

2.

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

and

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

•

3. By chain rule:

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

•

4. Substitute

u = 2 x

:

\frac{d y}{d x} = \frac{2}{1 - ( 2 x ) ^{2}} = \frac{2}{1 - 4 x ^{2}}

Answer: $\frac{d y}{d x} = \frac{2}{1 - 4 x ^{2}}$

Derivative of arccos x

Using a similar method for

y = arccos x

:

Start with

y = arccos x

, which means

x = cos y

Differentiate:

\frac{d x}{d y} = - sin y

Apply inverse formula:

\frac{d y}{d x} = \frac{1}{- s i n y} = \frac{- 1}{s i n y}

Use identity

sin y = 1 - cos^{2} y = 1 - x^{2}

Standard Result

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

Notice

The derivative of arccos is the negative of the derivative of arcsin. This makes sense because

arcsin x + arccos x = \frac{π}{2}

.

Derivative of arctan x

For

y = arctan x

:

Start with

y = arctan x

, which means

x = tan y

Differentiate:

\frac{d x}{d y} = sec^{2} y

Apply inverse formula:

\frac{d y}{d x} = \frac{1}{s e c ^{2} y} = cos^{2} y

Use identity

sec^{2} y = 1 + tan^{2} y = 1 + x^{2}

So

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

Standard Result

\frac{d}{d x} (arctan x) = \frac{1}{1 + x ^{2}}

Advantage

Unlike arcsin and arccos, arctan has no domain restrictions — it works for all real

x

.

Worked Example04

Differentiate $y = arctan (\frac{1 - x}{1 + x})$ .

•

1. Let

u = \frac{1 - x}{1 + x}

, so

y = arctan (u)

•

2. Find

\frac{d u}{d x}

using quotient rule:

•

\frac{d u}{d x} = \frac{( 1 + x ) ( - 1 ) - ( 1 - x ) ( 1 )}{( 1 + x ) ^{2}} = \frac{- 1 - x - 1 + x}{( 1 + x ) ^{2}} = \frac{- 2}{( 1 + x ) ^{2}}

•

3. Find

1 + u^{2}

:

•

1 + u^{2} = 1 + \frac{( 1 - x ) ^{2}}{( 1 + x ) ^{2}} = \frac{( 1 + x ) ^{2} + ( 1 - x ) ^{2}}{( 1 + x ) ^{2}} = \frac{2 + 2 x ^{2}}{( 1 + x ) ^{2}} = \frac{2 ( 1 + x ^{2} )}{( 1 + x ) ^{2}}

•

4. Apply chain rule:

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

•

5. Simplify:

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

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

Worked Example05

Differentiate $y = arcsin (x^{2})$ .

•

1. Use the chain rule: let

u = x^{2}

, so

y = arcsin (u)

•

2.

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

and

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

•

3. By chain rule:

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

•

4. Substitute

u = x^{2}

:

\frac{d y}{d x} = \frac{2 x}{1 - ( x ^{2} ) ^{2}} = \frac{2 x}{1 - x ^{4}}

Answer: $\frac{d y}{d x} = \frac{2 x}{1 - x ^{4}}$

Worked Example06

Differentiate $y = x arctan x$ .

•

1. Use the product rule:

y = u \cdot v

where

u = x

and

v = arctan x

•

2.

u^{'} = 1

and

v^{'} = \frac{1}{1 + x ^{2}}

•

3. Product rule:

\frac{d y}{d x} = u \cdot v^{'} + v \cdot u^{'}

•

4.

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

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

Worked Example07

Given $y = e^{x}$ , use the inverse function method to find $\frac{d}{d x} (ln x)$ .

•

1. If

y = e^{x}

, then the inverse is

x = ln y

•

2. We want

\frac{d x}{d y}

where

x = ln y

•

3. From

y = e^{x}

:

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

•

4. Using inverse formula:

\frac{d x}{d y} = \frac{1}{\frac{d y}{d x}} = \frac{1}{y}

•

5. So

\frac{d}{d y} (ln y) = \frac{1}{y}

Answer: $\frac{d}{d x} (ln x) = \frac{1}{x}$ (confirming the standard result)

Practice

Question 1: Differentiate

y = arccos (4 x)

.

•

1. Let

u = 4 x

, so

y = arccos (u)

•

2.

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

and

\frac{d u}{d x} = 4

•

3. Chain rule:

\frac{d y}{d x} = \frac{- 1}{1 - u ^{2}} \times 4 = \frac{- 4}{1 - 16 x ^{2}}

Answer: $\frac{d y}{d x} = \frac{- 4}{1 - 16 x ^{2}}$

Practice

Question 2: Differentiate

y = arctan (\frac{x}{2})

.

•

1. Let

u = \frac{x}{2}

, so

y = arctan (u)

•

2.

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

and

\frac{d u}{d x} = \frac{1}{2}

•

3. Chain rule:

\frac{d y}{d x} = \frac{1}{1 + u ^{2}} \times \frac{1}{2} = \frac{1}{2 ( 1 + \frac{x ^{2}}{4} )}

•

4. Simplify:

\frac{d y}{d x} = \frac{1}{2 + \frac{x ^{2}}{2}} = \frac{1}{\frac{4 + x ^{2}}{2}} = \frac{2}{4 + x ^{2}}

Answer: $\frac{d y}{d x} = \frac{2}{4 + x ^{2}}$

Practice

Question 3: Differentiate

y = (arcsin x)^{2}

.

•

1. Let

u = arcsin x

, so

y = u^{2}

•

2.

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

and

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

•

3. Chain rule:

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

Answer: $\frac{d y}{d x} = \frac{2 a r c s i n x}{1 - x ^{2}}$

Summary

Key Formulae

FunctionDerivative

y = arcsin x

\frac{1}{1 - x ^{2}}

y = arccos x

\frac{- 1}{1 - x ^{2}}

y = arctan x

\frac{1}{1 + x ^{2}}

Exam Tips

Always apply chain rule when the argument is not just

x

Remember the negative sign for arccos

arctan has no domain restrictions (works for all

x

)

These formulae are in the formula book — learn when to use them