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Derivatives of Inverse Functions

Last Updated : 13 May, 2024
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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. Or 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 inverse of function f. The inverse function is also known as the anti function. The inverse of function is denoted by  f-1.

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

Here, f and g are inverse functions.

Overview of Derivatives of Inverse Functions

Category

Subtopics

Key Points

Derivative of an Inverse Function

Inverse Function Theorem

This theorem is fundamental for deriving the derivatives of inverse functions, showing that the derivative of an inverse function is the reciprocal of the derivative of the original function at corresponding points.

Applying the Inverse Function Theorem

Practical application of the theorem to compute derivatives of inverse functions, illustrating step-by-step methodology.

Extending the Power Rule to Rational Exponents

Expands the power rule in calculus to handle cases with rational exponents, enhancing the toolkit for dealing with complex derivative problems.

Derivatives of Inverse Trigonometric Functions

Derivative of the Inverse Sine Function

Focuses on finding the derivative of arcsin(x), demonstrating its relationship with the derivative of the sine function.

Applying the Chain Rule to the Inverse Sine Function

Illustrates how to use the chain rule when differentiating composite functions that include the arcsin function, crucial for more complex calculations.

Applying Differentiation Formulas to an Inverse Tangent Function

Details methods to differentiate the arctan function using standard differentiation formulas, useful in trigonometric calculus.

Applying Differentiation Formulas to an Inverse Sine Function

Similar to the inverse tangent, this points out techniques for differentiating the arcsin function using established formulas.

Applying the Inverse Tangent Function

Explains how to apply the arctan function in calculus problems, particularly in solving integrals and derivatives involving arctan.

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) = ex

y = ex

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

x = ey

y = ln x

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

Inverse functions ex and ln x are symmetry about y = x.

Check: Derivative of Sin Inverse x

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.

Derivatives-of-Inverse-Functions

Derivatives of Inverse Functions

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) = ex, check whether condition holds true. Solution:

As, f(x) = ex

y = ex  

x = ln y

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

Now, 

f'(x) = \frac{d}{dx}      (ex) = ex

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}}      = ex

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’. 

xf(x)g(x)f'(x)
248\frac{-1}{6}
832\frac{1}{2}

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.

Check: Derivative of Inverse Trigonometric Functions

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:

1. 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}}\\

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}}\\

3. 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

4. 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

5. 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}}\\

6. 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}}\\

Examples:

Question 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}

Question 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)}}

Question 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)}}



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