Derivative of Inverse Trigonometric Functions

Inverse Trigonometric Functions

Inverse trigonometric functions - also called arcus functions, cyclometric functions, or anti-trigonometric functions - are used to find the angle corresponding to a given trigonometric value. In simple terms, if a trigonometric function gives you the ratio of sides of a triangle, its inverse tells you the angle that produces that ratio. These functions form the backbone of higher-level calculus, particularly when dealing with derivatives and integration.

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 play an important role in calculus, especially in solving differentiation problems involving trigonometric identities, integration, and coordinate geometry. These formulas help you find the rate of change of an angle when its trigonometric ratio changes with respect to a variable like $x$.

This section includes a table of standard derivatives, followed by step-by-step derivations for each inverse trigonometric function - $\sin^{-1}x$, $\cos^{-1}x$, $\tan^{-1}x$, $\cot^{-1}x$, $\sec^{-1}x$, and $\csc^{-1}x$.

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:

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

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

We have shared below the links to practice questions on the related topics of differentiation of inverse trigonometric functions: