Derivative of Inverse Trigonometric Functions

Inverse Trigonometric Functions is one of the important parts of Calculus, which applies to measuring the change in the function at a certain point. Mathematically, it forms a powerful tool by which slopes of functions are determined, the maximum and minimum of functions found, and problems on motion, growth, and decay, to name a few. These concepts of differentiation have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

In this article, we will cover the concept of the Inverse Trigonometric Functions. This concept falls under the broader category of Calculus, which is a crucial Chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last five years of the JEE Main exam (from 2013 to 2023), a total of two questions have been asked on this concept, including one in 2020, and one in 2021.

Differentiation of Inverse Trigonometric Function

These functions are also known as arcus functions, cyclometric functions, or anti-trigonometric functions. These functions are used to get an angle for a given trigonometric value. It refers to the change in the value of the trigonometric function at a certain rate. Inverse Trigonometric functions are the inverse functions sin, $\cos$, tan, etc.

14. $\frac{d}{d x}\left(\sin ^{-1}(\mathrm{x})\right)=\frac{1}{\sqrt{1-\mathrm{x}^2}}$

Let $\sin ^{-1} x=y$ where, $y \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

$
\begin{aligned}
& \therefore \quad \mathrm{x}=\sin \mathrm{y} \\
& \Rightarrow \quad \frac{d x}{d y}=\cos y=\sqrt{1-x^2} \quad(\because \cos y \geq 0 \quad \forall y \in[-\pi / 2, \pi / 2]) \\
& \Rightarrow \quad \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\sqrt{1-\mathrm{x}^2}}
\end{aligned}
$

15. $\frac{d}{d x}\left(\cos ^{-1}(\mathrm{x})\right)=-\frac{1}{\sqrt{1-\mathrm{x}^2}}$
16. $\frac{d}{d x}\left(\tan ^{-1}(\mathrm{x})\right)=\frac{1}{1+\mathrm{x}^2}$

Let $\tan ^{-1} x=y$ where, $y \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

$
\begin{aligned}
\therefore & x & =\tan y \\
\Rightarrow & \frac{d x}{d y} & =\sec ^2 y \\
\Rightarrow & & =1+\tan ^2 y \\
\Rightarrow & & =1+x^2 \\
\Rightarrow & \frac{d y}{d x} & =\frac{1}{1+x^2}
\end{aligned}
$

17. $\frac{d}{d x}\left(\cot ^{-1}(\mathrm{x})\right)=-\frac{1}{1+\mathrm{x}^2}$

As

$
\cot ^{-1}(x)=\frac{\pi}{2}-\tan ^{-1}(x)
$

Differentiating both sides we get,

$
\frac{d\left(\cot ^{-1}(x)\right)}{d x}=-\frac{1}{1+x^2}
$
18. $\quad \frac{d}{d x}\left(\sec ^{-1}(\mathbf{x})\right)=\frac{1}{|\mathbf{x}| \sqrt{\mathbf{x}^2-1}}$
19. $\frac{d}{d x}\left(\csc ^{-1}(\mathbf{x})\right)=-\frac{1}{|\mathbf{x}| \sqrt{\mathbf{x}^2-1}}$

Substitutions for Inverse trigonometric functions

\begin{array}{ccc}
\hline \text { S. No. } & \text { Expression } & \text { Substitutions } \\
\hline \hline 1 . & \sqrt{a^2-x^2} & x=a \sin \theta \text { or } a \cos \theta \\
\hline 2 . & \sqrt{a^2+x^2} & x=a \tan \theta \text { or } a \cot \theta \\
\hline 3 . & \sqrt{x^2-a^2} & x=a \sec \theta \text { or } a \csc \theta \\
\hline 4 . & \sqrt{\frac{a+x}{a-x}} \text { or } \sqrt{\frac{a-x}{a+x}} & x=a \cos \theta \text { or } a \cos 2 \theta \\
\hline 5 . & \sqrt{a+x} \text { or } \sqrt{a-x} & x=a \cos \theta \text { or } a \cos 2 \theta
\end{array}

The domain of inverse functions is:
1. $\sin ^{-1} x[-1,1]$
2. $\cos ^{-1} x[-1,1]$
3. $\tan ^{-1} \times R$
4. $\operatorname{cosec}^{-1} x(-\infty,-1] \cup[1, \infty)$
5. $\sec ^{-1} x(-\infty,-1] \cup[1, \infty)$
6. $\cot ^{-1} \times \mathrm{R}$

Principal Value of function f^-1 (f (x))

1. $\sin ^{-1}(\sin (\theta))=\theta \quad$ for all $\theta \in[-\pi / 2, \pi / 2]$
2. $\cos ^{-1}(\cos (\theta))=\theta \quad$ for all $\theta \in[0, \pi]$
3. $\tan ^{-1}(\tan (\theta))=\theta \quad$ for all $\theta \in(-\pi / 2, \pi / 2)$
4. $\cot ^{-1}(\cot (\theta))=\theta \quad$ for all $\theta \in(0, \pi)$
5. $\sec ^{-1}(\sec (\theta))=\theta \quad$ for all $\theta \in[0, \pi]-\{\pi / 2\}$
6. $\csc ^{-1}(\csc (\theta))=\theta \quad$ for all $\theta \in[-\pi / 2, \pi / 2]-\{0\}$

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.

Summary

Inverse trigonometric function is an important part of calculus. Inverse Trig Derivative is the rate of change in the inverse trigonometric functions concerning independent variables. It provides a deeper understanding of mathematical ideas paramount for later developments in many scientific and engineering disciplines.