Derivative of Inverse Trigonometric Functions

Inverse Trigonometric Functions

Inverse trigonometric functions, also known as anti-trigonometric functions, arcus functions, or cyclometric functions, are used to determine the angle corresponding to a given trigonometric value. In simple terms, while a trigonometric function gives the ratio of sides of a triangle, its inverse function helps you find the angle that produces that ratio.

These functions play a crucial role in Class 11 calculus, especially in topics related to differentiation and integration. A strong understanding of inverse trigonometric functions is essential for solving higher-order calculus problems, simplifying expressions, and handling questions commonly asked in board exams and competitive entrance exams.

Inverse trigonometric functions are the inverse of standard trigonometric functions like $\sin$, $\cos$, $\tan$, $\cot$, $\sec$, and $\csc$.

Domains of Inverse Trigonometric Functions

The domain defines the set of all possible input values for which the inverse trigonometric functions are defined:

1. $\sin^{-1}x$, domain: $[-1, 1]$

2. $\cos^{-1}x$, domain: $[-1, 1]$

3. $\tan^{-1}x$, domain: $\mathbb{R}$

4. $\csc^{-1}x$, domain: $(-\infty, -1] \cup [1, \infty)$

5. $\sec^{-1}x$, domain: $(-\infty, -1] \cup [1, \infty)$

6. $\cot^{-1}x$, domain: $\mathbb{R}$

Principal Values of Inverse Trigonometric Functions

The principal value defines the main range for which the inverse trigonometric function gives a unique value.

1. $\sin^{-1}(\sin \theta) = \theta$, for all $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

2. $\cos^{-1}(\cos \theta) = \theta$, for all $\theta \in [0, \pi]$

3. $\tan^{-1}(\tan \theta) = \theta$, for all $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

4. $\cot^{-1}(\cot \theta) = \theta$, for all $\theta \in (0, \pi)$

5. $\sec^{-1}(\sec \theta) = \theta$, for all $\theta \in [0, \pi] - {\frac{\pi}{2}}$

6. $\csc^{-1}(\csc \theta) = \theta$, for all $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - {0}$

Range and Restrictions for Each Inverse Function

Each inverse trigonometric function comes with restrictions to ensure it gives a unique output angle for every valid input value of $x$.

For instance:

• $\sin^{-1}(x)$ and $\tan^{-1}(x)$ are odd functions, meaning $\sin^{-1}(-x) = -\sin^{-1}(x)$ and $\tan^{-1}(-x) = -\tan^{-1}(x)$.

• $\cos^{-1}(x)$, on the other hand, is neither odd nor even, because it doesn’t show symmetry about the origin or the y-axis.

• $\sec^{-1}(x)$ and $\csc^{-1}(x)$ are defined only for $|x| \geq 1$, since their corresponding trigonometric values never lie between $-1$ and $1$.

Principal Values, Domains, and Ranges of Inverse Trigonometric Functions

To define an inverse function properly, each trigonometric function is restricted to a specific principal branch where it is one-to-one. The corresponding inverse is then defined on that range.

Differentiation of Inverse Trigonometric Functions

The derivatives of inverse trigonometric functions are a core concept in Class 11 calculus and are widely used in solving differentiation problems, simplifying trigonometric identities, evaluating integrals, and handling questions from coordinate geometry. These derivative formulas help determine the rate of change of an inverse trigonometric function with respect to a variable like $x$, which is essential for both board exams and competitive exams.

This section presents a comprehensive table of standard derivatives of inverse trigonometric functions, followed by clear, step-by-step derivations of each formula. The inverse functions covered include $\sin^{-1}x$, $\cos^{-1}x$, $\tan^{-1}x$, $\cot^{-1}x$, $\sec^{-1}x$, and $\csc^{-1}x$, explained using simple methods aligned with the Class 11 NCERT syllabus.

Step-by-Step Differentiation Rules

The derivation of these formulas is based on the chain rule and implicit differentiation. To find the derivative of an inverse trigonometric function, assume the function equals an angle (say $y$), then convert it into its trigonometric form and differentiate both sides with respect to $x$.

Let’s derive each one below.

Derivative of $\sin^{-1}x$

Let $y = \sin^{-1}x$
Then, $\sin y = x$

Differentiating both sides with respect to $x$:

$\cos y \dfrac{dy}{dx} = 1$

We know $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$

So, $\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - x^2}}$

Hence,
$\dfrac{d}{dx}(\sin^{-1}x) = \dfrac{1}{\sqrt{1 - x^2}}$

Derivative of $\cos^{-1}x$

Let $y = \cos^{-1}x$
Then, $\cos y = x$

Differentiating both sides with respect to $x$:

$-\sin y \dfrac{dy}{dx} = 1$

We know $\sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - x^2}$

So, $\dfrac{dy}{dx} = -\dfrac{1}{\sqrt{1 - x^2}}$

Therefore,
$\dfrac{d}{dx}(\cos^{-1}x) = -\dfrac{1}{\sqrt{1 - x^2}}$

Derivative of $\tan^{-1}x$

Let $y = \tan^{-1}x$
Then, $\tan y = x$

Differentiating both sides with respect to $x$:

$\sec^2 y \dfrac{dy}{dx} = 1$

We know $\sec^2 y = 1 + \tan^2 y = 1 + x^2$

So, $\dfrac{dy}{dx} = \dfrac{1}{1 + x^2}$

Hence,
$\dfrac{d}{dx}(\tan^{-1}x) = \dfrac{1}{1 + x^2}$

Derivative of $\cot^{-1}x$

Let $y = \cot^{-1}x$
Then, $\cot y = x$

Differentiating both sides with respect to $x$:

$-\csc^2 y \dfrac{dy}{dx} = 1$

We know $\csc^2 y = 1 + \cot^2 y = 1 + x^2$

So, $\dfrac{dy}{dx} = -\dfrac{1}{1 + x^2}$

Therefore,
$\dfrac{d}{dx}(\cot^{-1}x) = -\dfrac{1}{1 + x^2}$

Derivative of $\sec^{-1}x$

Let $y = \sec^{-1}x$
Then, $\sec y = x$

Differentiating both sides with respect to $x$:

$\sec y \tan y \dfrac{dy}{dx} = 1$

So, $\dfrac{dy}{dx} = \dfrac{1}{\sec y \tan y}$

Now,
$\sec y = x$ and $\tan y = \sqrt{x^2 - 1}$

Hence, $\dfrac{dy}{dx} = \dfrac{1}{x\sqrt{x^2 - 1}}$

For all valid $x$, we take the absolute value of $x$, giving:

$\dfrac{d}{dx}(\sec^{-1}x) = \dfrac{1}{|x|\sqrt{x^2 - 1}}$

Derivative of $\csc^{-1}x$

Let $y = \csc^{-1}x$
Then, $\csc y = x$

Differentiating both sides with respect to $x$:

$-\csc y \cot y \dfrac{dy}{dx} = 1$

So, $\dfrac{dy}{dx} = -\dfrac{1}{\csc y \cot y}$

We know $\csc y = x$ and $\cot y = \sqrt{x^2 - 1}$

Therefore, $\dfrac{dy}{dx} = -\dfrac{1}{x\sqrt{x^2 - 1}}$

Including the absolute value of $x$ for general validity:

$\dfrac{d}{dx}(\csc^{-1}x) = -\dfrac{1}{|x|\sqrt{x^2 - 1}}$

Important Identities Used in Differentiating Inverse Trigonometric Functions

Below is the table of essential identities that commonly appear while differentiating inverse trigonometric functions. These identities help simplify expressions, manage square roots, and apply the correct derivative formulas with ease.

Solved Examples Based on Rules of Inverse Trigonometric Function

Example 1: The derivative of $\tan ^{-1}\left\{\frac{\sqrt{1+x^2}-1}{x}\right\}$ with respect to $\tan ^{-1}\left\{\frac{2 x \sqrt{1-x^2}-1}{1-2 x^2}\right\}$ at $x=\frac{1}{2}$ is:
[JEE Main 2020]
1) $\frac{2 \sqrt{3}}{5}$
2) $\frac{\sqrt{3}}{10}$
3) $\frac{2 \sqrt{3}}{3}$
4) $\frac{\sqrt{3}}{12}$

Solution:

$\begin{aligned} & \tan ^{-1}\left\{\frac{\sqrt{1+x^2}-1}{x}\right\} \\ & \text { Let } x=\tan \theta \\ & \tan ^{-1}\left\{\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right\}=\frac{\theta}{2}=\frac{\tan ^{-1} x}{2} \\ & \tan ^{-1}\left\{\frac{2 x \sqrt{1-x^2}}{1-2 x^2}\right\} \quad \text { let } x=\sin \alpha \\ & =\tan ^{-1}\left\{\frac{2 \sin \alpha \sqrt{1-\sin ^2 \alpha}}{1-2 \sin ^2 \alpha}\right\} \\ & =2 \alpha=2 \sin ^{-1} x\end{aligned}$

$\begin{aligned} & \frac{d y}{d x}=\frac{\frac{1}{2}+\frac{1}{1+x^2}}{2 \times \frac{1}{\sqrt{1-x^2}}} \\ & =\frac{1}{4} \frac{\sqrt{1-x^2}}{1+x^2} \\ & x=\frac{1}{2} \\ & \text { at } \\ & \frac{1}{4} \times \frac{\sqrt{1-\frac{1}{4}}}{1+\frac{1}{4}}=\frac{\frac{\sqrt{3}}{2}}{5}=\frac{\sqrt{3}}{10}\end{aligned}$

Hence, the answer is the option 2.

Example 2: If $y(x)=\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right), x \in\left(\frac{\pi}{2}, \pi\right)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=\frac{5 \pi}{6}$ is :
[JEE Main 2021]
1) $-\frac{1}{2}$
2) $-1$
3) $0$
4) $\frac{1}{2}$

Solution:

$\begin{aligned} x & \in\left(\frac{\pi}{4}, \pi\right) \\ \Rightarrow & \frac{x}{2} \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \\ y & =\left(\frac{\sqrt{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^2}+\sqrt{\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)^2}}{\sqrt{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^2}-\sqrt{\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)^2}}\right) \\ & =\cot ^{-1}\left(\frac{2 \sin \frac{x}{2}}{2 \cos \frac{x}{2}}\right)=\cot ^{-1} \tan \frac{x}{2} \\ & =\frac{\pi}{2}-\tan ^{-1} \tan \frac{x}{2}=\frac{\pi}{2}-\frac{x}{2} \\ \frac{d y}{d x} & =\frac{-1}{2}\end{aligned}$

Hence, the answer is the option 1.

Example 3: If $x=2^{\sin ^{-1} t}$ and $y=2^{\cos ^{-1} t}(|t| \leq 1)$, then $\frac{d y}{d x}$ is equal to
1) $y / x$
2) $x / y$
3) $-y / x$
4) $1$

Solution:

$
\begin{aligned}
& x=2^{\sin ^{-1} t} \\
& \text { So } \frac{d x}{d t}=2^{\sin ^{-1} t} \cdot \ln (2) \cdot \frac{1}{\sqrt{1-x^2}} \\
& y=2^{\cos ^{-1} t} \\
& \text { So } \frac{d y}{d t}=2^{\cos ^{-1} t} \cdot \ln (2) \cdot \frac{-1}{\sqrt{1-x^2}} \\
& \therefore \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2^{\cos ^{-1} t} \cdot \ln (2) \cdot \frac{-1}{\sqrt{1-x^2}}}{2^{\sin -1} t \cdot \ln (2) \cdot \frac{1}{\sqrt{1-x^2}}}=\frac{-2^{\cos ^{-1} t}}{2^{\sin ^{-1} t}}=-\frac{y}{x}
\end{aligned}
$

Example 4: Let $f(x)=\sec ^{-1}\left(1+x^2\right)$, then $\mathrm{f}(\mathrm{x})$ equals
1) $\frac{2 x}{\left(x^2+1\right) \sqrt{x^4+2 x^2}}$
2) $\frac{2 x}{\left(x^2+1\right) \sqrt{x^2+2}}$
3) $\frac{1}{\left(x^2+1\right) \sqrt{x^4+2 x^2}}$
4) $\frac{x}{\left(x^2+1\right) \sqrt{x^4+2 x^2}}$

Solution:
Let $\left(1+x^2\right)=u$
So $y=\sec ^{-1}(u)$

Using Chain Rule

$\begin{aligned} & \frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x} \\ & \frac{d y}{d x}=\frac{d\left(\sec ^{-1}(u)\right)}{d u} \cdot \frac{d\left(1+x^2\right)}{d x} \\ & \frac{d y}{d x}=\frac{1}{|u| \sqrt{u^2-1}} \cdot 2 x \\ & \frac{d y}{d x}=\left(\frac{1}{\left(x^2+1\right) \sqrt{\left(x^2+1\right)^2-1}}\right) \cdot 2 x \\ & \frac{d y}{d x}=\frac{2 x}{\left(x^2+1\right) \sqrt{\left(x^4+2 x^2\right)}}\end{aligned}$

Hence, the answer is the option 1.

Example 5: $\frac{d}{d x}\left(\tan ^{-1} \sqrt{\frac{1-\cos (x)}{1+\cos (x)}}\right)$ is equal to (Given $x \in(0, \pi)$ )
1) $-\frac{1}{4}$
2) $\frac{1}{2}$
3) $-\frac{1}{2}$
4) $\frac{1}{4}$

Solution:

In many questions involving differentiation of complex inverse trigonometric functions, we need to first simplify the given expression using techniques learned in the concept 'Simplification using Substitution' in the chapter 'Inverse Trigonometric Functions'.

Let $y=\tan ^{-1} \sqrt{\frac{1-\cos (x)}{1+\cos (x)}}$ $=\tan ^{-1} \sqrt{\frac{2 \sin ^2 \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}}$
$=\tan ^{-1}\left|\tan \frac{x}{2}\right|$ As $x \in(0, \pi)$, so $\frac{x}{2} \in\left(0, \frac{\pi}{2}\right)$, and $\tan \left(\frac{x}{2}\right)$ is positive

$
\begin{aligned}
& =\tan ^{-1} \tan \left(\frac{x}{2}\right)=\frac{x}{2} \\
& \therefore \frac{d y}{d x}=\frac{1}{2}
\end{aligned}
$

Hence, the answer is the option 2.

List of topics related to the Differentiation of Inverse Trigonometric Functions

This section covers all key topics related to the differentiation of trigonometric functions such as $\sin x$, $\cos x$, $\tan x$, and their inverses. You’ll find a complete list of subtopics to strengthen your calculus foundation for exams like JEE and CBSE Class 12.

Differentiability and Existence of Derivative

Examining differentiability Using Graph of Function

Continuity and Discontinuity

Continuity of Composite Function

Continuity And Differentiability

Differentiability of Composite Function

NCERT Resources

Here, you’ll get direct access to NCERT exemplar questions, step-by-step solutions, and detailed notes based on differentiation of trigonometric functions - ideal for board exam preparation and concept clarity.

NCERT Class 12 Maths Notes for Chapter 5 - Continuity and Differentiability

NCERT Class 12 Maths Solutions for Chapter 5 - Continuity and Differentiability

NCERT Class 12 Maths Exemplar Solutions for Chapter 5 - Continuity and Differentiability

Practice Questions based on Differentiation of Inverse Trigonometric Functions

This section includes chapter-wise and mixed-level practice questions covering standard derivatives, higher-order derivatives, and real-life applications of trigonometric differentiation to help you master problem-solving skills.

Differentiation Of Inverse Trigonometric Function - Practice Question MCQ

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