Derivatives of Inverse Functions

In mathematics, a function(e.g. f), is said to be an inverse of another(e.g. g), if given the output of g returns the input value given to f. Additionally, this must hold true for every element in the domain co-domain(range) of g. E.g. assuming x and y are constants if g(x) = y and f(y) = x then the function f is said to be an inverse of the function g. In other words, if a function f: A ⇢ B is one-one and onto function or bijective function, then a function defined by g: B ⇢ A is known as the inverse of function f. The inverse function is also known as the anti-function. The inverse of the function is denoted by  f^-1.

f(g(x)) = g(f(x)) = x

Here, f and g are inverse functions.

What are Derivatives of Inverse Functions?

The derivative of an inverse function provides a way to find the rate of change of an inverse function at a point. If you have a function y = f(x) and its inverse x=f⁻¹(y), the derivatives of these functions are related in a specific way. The details of the derivative of inverse functions have been tabulated below:

Procedure of Finding Inverse of f

• Check for one-one and onto function.
• If invertible then, interchange x and y in definition of f(x).
• Find y in terms of x.
• Obtained y is inverse of f defined from B ⇢ A.

Example: f(x) = e^x

y = e^x

Inverse of f(x) will be obtained by y ↔ x

x = e^y

y = ln x

Now, e^x and ln x are inverse function of each other.

Check: Derivative of Sin Inverse x

Relation between Derivatives of Inverse Functions

As we already know that what is inverse function, so now we are going to find the derivative of inverse function. So, if a function f(x) is a continuous one-to-one function or bijective function defined on an interval lets say I, then its inverse is also continuous and if the function f(x) is a differentiable function, then its inverse is also a differentiable function.

g'(x) = \mathbf{\frac{1}{f'(g(x))}} 

Here, f and g are inverse function. It is known as inverse function theorem.

Proof:

Let us considered f and g be the inverse functions and x is present in the g domain, then 

f(g(x)) = x

Differentiate both side with respect to x 

\frac{d f(g(x))}{x} = \frac{d(x)}{dx}

Now solve the LHS using chain rule we get

f'(g(x))g'(x) = 1

Now solve for g'(x):

g'(x) = {\frac{1}{f'(g(x))}} 

Hence, the derivative on inverse function is solved.

Example 1: f(x) = e^x, check whether condition holds true.

Solution:

As, f(x) = ex

y = e^x  

x = ln y

g(x) = f^-1(x) = ln x

Now, 

f'(x) = \frac{d}{dx}      (e^x) = e^x

g'(x) = \frac{d}{dx}      (ln x) = \frac{1}{x}

g'(f(x)) = \frac{1}{e^x}

\frac{1}{g'(f(x))} = \frac{1}{\frac{1}{e^x}}      = e^x

Hence, 

f'(x) = \frac{1}{g'(f(x))}     , holds true.

Example 2: Let f(x) = \frac{1}{2}     x³ + 3x - 4, and let g be the inverse function of f where f(-2) = -14. Find g'(-14)

Solution:

f'(x) = \frac{1}{2}      (3x²) + 3

According to eq (1).

g'(x) = \mathbf{\frac{1}{f'(g(x))}}

As, f(x) = g^-1(x) and f(-2) = -14

then g(-14) = -2

x = -14

g'(-14) = \frac{1}{f'(g(-14))}

g'(-14) = \frac{1}{f'(-2)}

f'(-2) = \frac{1}{2}      (3(-2)²) + 3

f'(-2) = \frac{12}{2}     + 3

f'(-2) = 9

then, g'(-14) = \frac{1}{9}

How to Find Derivatives of Inverse Functions from the Table?

Let us discuss this concept with the help of an example. So let us assume g and f be the inverse function and the following table lists a few values of f, g and f'. 

We have to find g'(2). As it said from the question, that f and g be inverse functions. This means if we have two sets, now let us assume that the first set is the domain of f. So in this set, if we start from some x value then f is going to map that x to another value which is known as f(x)(this is the use of function f). Now as we know that g is the inverse of f, so this g gets us back to the first set(this is the use of function g).

Hence we get 

g(f(x)) = x ...(i)

f(g(x)) = x ...(ii)

Where, both are valid.

From eq(ii), we have

f(g(x)) = x 

Now differentiate both side w.r.t. x. we get

\frac{d(f(g(x)))}{dx} = \frac{d(x)}{dx}

Now on the LHS we apply chain rule, now we get

f'(g(x))g'(x) = 1

g'(x) = \mathbf{\frac{1}{f'(g(x))}} 

Now we are going to find the value of g'(2)

g'(2) = \frac{1}{f'(g(2))}

From the table we get the value of g(2)

g'(2) = \frac{1}{f'(8)}

From the table we get the value of f'(8)

g'(2) = \frac{1}{\frac{1}{2}}

g'(2) = 2

Hence, the value of g'(2) = 2.

Derivatives of Inverse Trigonometric Functions

Now, let's focus on figuring out the derivatives of inverse trigonometric functions. These derivatives will be super helpful when we delve into integration later on. What's interesting is that the derivatives of inverse trigonometric functions are actually algebraic functions, which might come as a surprise.

How to find the derivatives of inverse trigonometric functions?

We remark that inverse trigonometric functions are continuous functions. Now we use first principles and chain rule to find derivatives of these functions:

Derivative of f given by f(x) = sin^–1 x.

From first principle 

f(x) = sin^–1 x and f(x+h) = sin^–1 (x+h)

\frac{d}{dx}(f(x)) = \lim_{h\to0} \frac{f(x+h)-f(x)}{h} \\ = \lim_{h\to0} \frac{sin^{-1}(x+h)-sin^{-1}x}{h}\\ = \lim_{h\to0} \frac{sin^{-1}[(x+h)\sqrt{1-x^2} - x\sqrt{1-(x+h)^2}]}{h}

Using the formula,

sin^–1 x - sin^–1 y = sin-1(x \sqrt{1-y^2}        - y \sqrt{1-x^2}       )

= \lim_{h\to0} \frac{sin^{-1}[(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}]}{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}} \times \frac{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}}{h}

As

\lim_{x\to0} (\frac{sin^{-1}x}{x}) = 1 \\ So, \\ = \lim_{h\to0} 1. \frac{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}}{h} \\ = \lim_{h\to0} \frac{(x+h)^2(1-x^2)-x^2(1-(x+h)^2)}{h} \times \frac{1}{(x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}}\\ = \lim_{h\to0} \frac{(x+h)^2-x^2}{h} \times \frac{1}{(x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}}\\ = \lim_{h\to0} (2x+h) \times \frac{1}{(x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}}\\ = \frac{(2x)}{2x \sqrt{1-x^2}}\\ = \frac{1}{\sqrt{1-x^2}} 

Hence

\frac{d}{dx} (sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}

From chain rule 

y = sin^-1x

sin y = x

Differentiating both sides w.r.t x, we get

\frac{d}{dx}(sin \hspace{0.1cm}y) = 1 \\ \frac{d}{dx}(sin \hspace{0.1cm}y) \times \frac{dy}{dx}  = 1\\ \cos\hspace{0.1cm}y\frac{dy}{dx}  = 1\\ \frac{dy}{dx}  = \frac{1}{cos\hspace{0.1cm}y}\\ \frac{dy}{dx}  = \frac{1}{\sqrt{1-sin^2\hspace{0.1cm}y}}

As

\sin\hspace{0.1cm}y = x \\ \frac{dy}{dx}  = \frac{1}{\sqrt{1-x^2}}\\

Derivative of f given by f(x) = cos^–1 x.

From first principle

f(x) = cos^–1 x and f(x+h) = cos^–1 (x+h)

\frac{d}{dx}(f(x)) = \lim_{h\to0} \frac{f(x+h)-f(x)}{h} \\ = \lim_{h\to0} \frac{cos^{-1}(x+h)-cos^{-1}x}{h}\\ = \lim_{h\to0} \frac{(\frac{\pi}{2}-sin^{-1}(x+h))-(\frac{\pi}{2}-sin^{-1}x)}{h}

As

cos^{-1} x = \frac{\pi}{2} - sin^{-1} x \\ = - \lim_{h\to0} \frac{sin^{-1}(x+h)-sin^{-1}x)}{h}

By using the previous result

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

Hence, 

\frac{d}{dx} (cos^{-1}x) = -\frac{1}{\sqrt{1-x^2}}

From chain rule

y = cos^-1x

cos y = x

Differentiating both sides w.r.t x, we get

\frac{d}{dx}(cos \hspace{0.1cm}y) = 1 \\ \frac{d}{dx}(cos \hspace{0.1cm}y) \times \frac{dy}{dx}  = 1\\ -sin\hspace{0.1cm}y\frac{dy}{dx}  = 1\\ \frac{dy}{dx}  = - \frac{1}{sin\hspace{0.1cm}y}\\ \frac{dy}{dx}  = -\frac{1}{\sqrt{1-cos^2\hspace{0.1cm}y}}

As

cos\hspace{0.1cm}y = x \\ \frac{dy}{dx}  = -\frac{1}{\sqrt{1-x^2}}\\

Derivative of f given by f(x) = tan^–1 x.

From first principle

f(x) = tan^–1 x and f(x+h) = tan^–1 (x+h)

\frac{d}{dx}(f(x)) = \lim_{h\to0} \frac{f(x+h)-f(x)}{h} \\ = \lim_{h\to0} \frac{tan^{-1}(x+h)-tan^{-1}x}{h}

As

tan^{-1} x - tan^{-1}y = tan^{-1}(\frac{x-y}{1+xy}) \\ = \lim_{h\to0} \frac{tan^{-1}(\frac{x+h-x}{1+x(x+h)})}{h}\\ = \lim_{h\to0} \frac{tan^{-1}(\frac{h}{1+x^2+xh})}{\frac{h}{1+x^2+xh}} \times \frac{1}{1+x^2+xh}\\ = 1. \frac{1}{1+x^2}

Hence

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

From chain rule

y = tan^-1x

tan y = x

Differentiating both sides w.r.t x, we get

\frac{d}{dx}(tan \hspace{0.1cm}y) = 1 \\ \frac{d}{dx}(tan \hspace{0.1cm}y) \times \frac{dy}{dx}  = 1\\ sec^2\hspace{0.1cm}y\frac{dy}{dx}  = 1\\ \frac{dy}{dx}  = \frac{1}{sec^2\hspace{0.1cm}y}\\ \frac{dy}{dx}  = \frac{1}{1+tan^2\hspace{0.1cm}y}

As

tan y = x dy/dx = 1/1+x²

Derivative of f given by f(x) = cot^–1 x.

From first principle

f(x) = cot^–1 x and f(x+h) = cot^–1 (x+h)

\frac{d}{dx}(f(x)) = \lim_{h\to0} \frac{f(x+h)-f(x)}{h} \\ = \lim_{h\to0} \frac{cot^{-1}(x+h)-cot^{-1}x}{h}

As

cot^{-1} x = \frac{\pi}{2} - tan^{-1}x\\ = \lim_{h\to0} \frac{(\frac{\pi}{2} - tan^{-1}(x+h))-(\frac{\pi}{2} - tan^{-1}x)}{h}\\ = - \lim_{h\to0} \frac{tan^{-1}(x+h)-tan^{-1}x}{h}

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

Hence

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

From chain rule

y = cot^-1x

cot y = x

Differentiating both sides w.r.t x, we get

\frac{d}{dx}(cot \hspace{0.1cm}y) = 1 \\ \frac{d}{dx}(cot \hspace{0.1cm}y) \times \frac{dy}{dx}  = 1\\ -cosec^2\hspace{0.1cm}y\frac{dy}{dx}  = 1\\ \frac{dy}{dx}  = \frac{-1}{cosec^2\hspace{0.1cm}y}\\ \frac{dy}{dx}  = \frac{-1}{1+cot^2\hspace{0.1cm}y}

As

cot y = x dy/dx = -1/1 + x²

Derivative of f given by f(x) = sec^–1 x.

From chain rule

y = sec^-1x

sec y = x

Differentiating both sides w.r.t x, we get

\frac{d}{dx}(sec \hspace{0.1cm}y) = 1 \\ \frac{d}{dx}(sec \hspace{0.1cm}y) \times \frac{dy}{dx}  = 1\\ sec\hspace{0.1cm}y \hspace{0.1cm}tan\hspace{0.1cm}y\frac{dy}{dx}  = 1\\ \frac{dy}{dx}  = \frac{1}{sec\hspace{0.1cm}y \hspace{0.1cm}tan\hspace{0.1cm}y}\\ \frac{dy}{dx}  = \frac{1}{|sec\hspace{0.1cm}y| \hspace{0.1cm}|tan\hspace{0.1cm}y|}\\

If x > 1, then y ∈ (0, π/2)

∴ sec y > 0, tan y > 0 ⇒ |sec y||tan y| = sec y tan y

If x < -1, then y ∈ (π/2,π)

∴ sec y < 0, tan y < 0 ⇒ |sec y||tan y| = (-sec y) (-tan y) = sec y tan y

\frac{dy}{dx} = \frac{1}{|sec \hspace{0.1cm}y| \sqrt{tan^2y}}\\ \frac{dy}{dx} = \frac{1}{|sec \hspace{0.1cm}y| \sqrt{1-sec^2y}}

As

sec y = x 

\frac{dy}{dx}  = \frac{1}{|x|\sqrt{x^2-1}}\\

Derivative of f given by f(x) = cosec^–1 x.

From chain rule

y = cosec^-1x

cosec y = x

Differentiating both sides w.r.t x, we get

\frac{d}{dx}(cosec \hspace{0.1cm}y) = 1 \\ \frac{d}{dx}(cosec \hspace{0.1cm}y) \times \frac{dy}{dx}  = 1\\ -cosec\hspace{0.1cm}y \hspace{0.1cm}cot\hspace{0.1cm}y\frac{dy}{dx}  = 1\\ \frac{dy}{dx}  = \frac{1}{-cosec\hspace{0.1cm}y \hspace{0.1cm}cot\hspace{0.1cm}y}\\ \frac{dy}{dx}  = \frac{-1}{|cosec\hspace{0.1cm}y| \hspace{0.1cm}|cot\hspace{0.1cm}y|}\\

If x > 1, then y ∈ (0, π/2)

∴ cosec y > 0, cot y > 0 ⇒ |cosec y||cot y| = cosec y cot y

If x < -1, then y∈ (-π/2, 0)

∴ cosec y < 0, cot y < 0 ⇒ |cosec y||cot y| = (-cosec y) (-cot y)

\frac{dy}{dx} = \frac{-1}{|cosec \hspace{0.1cm}y| \sqrt{cot^2y}}\\ \frac{dy}{dx} = \frac{-1}{|cosec \hspace{0.1cm}y| \sqrt{cosec^2y-1}}

As

cosec y = x

\frac{dy}{dx}  = \frac{-1}{|x|\sqrt{x^2-1}}\\

Also, Check

• Differentiation of Inverse Trigonometric Functions
• Derivative Rules
• Derivative Formulas

Solved Example on Derivative of Inverse Function

Example 1. Find the derivative of y = tan ^-1 (x²).

Solution:

Differentiating both side, we get

\frac{dy}{dx} = \frac{d(tan^{-1} (x^2))}{dx}      

Using the inverse derivative of tan^-1 θ

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

 = \frac{2x}{1+x^4}

Example 2. Find the derivative of y = sin ^-1 (3x-2).

Solution:

Differentiating both side, we get

\frac{dy}{dx} = \frac{d (sin^{ -1} (3x-2))}{dx}   

Using the inverse derivative of sin^-1 θ

= \frac{1}{\sqrt{1-(3x-2)^2}} \frac{d(3x-2)}{dx}      

= \frac{1}{\sqrt{1-(9x^2-18x+4)}}  (3)

= \frac{3}{\sqrt{(18x-9x^2-4)}}

Example 3. Find the derivative of y = cos ^-1 (1 - x²).

Solution:

Differentiating both side, we get

\frac{dy}{dx} = \frac{d(cos^{ -1} (1 - x^2))}{dx}      

Using the inverse derivative of cos^-1 θ

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

= \frac{-1}{\sqrt{1-(1-2x^2+x^4)}}  (-2x)

= \frac{2x}{\sqrt{(2x^2-x^4)}}

= \frac{2}{\sqrt{(2-x^2)}}

Practice Problems - Derivatives of Inverse Functions

1: Find the inverse function:f(x)=3x−5

2: Find the inverse function: f(x)=\frac{2x + 1}{3}

3: Find the derivative of the inverse function at y=4 if:f(x)=x²+1

4: Find the inverse function:f(x)= \frac{1}{x}

5: Find the derivative of the inverse function at y=2 if:f(x)=ln⁡(x)

6: Find the inverse function:f(x)=x³+2

7: Find the derivative of the inverse function at y=5 if:f(x)=e^x

8: Find the inverse function: f(x) = \sqrt{x - 1}

Summary

The derivatives of inverse functions are found using a specific formula that relates the derivative of a function to the derivative of its inverse. If f(x) is a function with an inverse f⁻¹(y), and fff is differentiable with a non-zero derivative, then the derivative of its inverse at a point y is given by \frac{d}{dy}[f^{-1}(y)] = \frac{1}{f'(x)}, where x=f⁻¹(y). This formula is derived from the fact that if y=f(x), then x=f⁻¹(y), and differentiating both sides with respect to yyy leads to \frac{dx}{dy} \cdot \frac{dy}{dx} = 1. Hence, \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}​. This relationship is crucial in finding the derivatives of inverse functions without explicitly solving for the inverse function.

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Article Tags :

• Mathematics
• School Learning
• Class 12
• Maths-Class-12
• Calculus

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