Question

$\frac{d}{dx}\left({\tan }^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)\right)=$

• A. $\frac{a}{\left(a^2+x^2\right)}$
• B. $\frac{1}{\sqrt{a^2-x^2}}$
• C. $-\frac{1}{\left(a^2+x^2\right)}$
• D. none of these

Answer 1

The correct option is B

$\frac{1}{\sqrt{a^2-x^2}}$

Explanation for correct option:

Step-1: Simplify the given data.

Given, $\frac{d}{dx}\left({\tan }^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)\right)$

Put, $x=a\sin \left(\theta \right)\Rightarrow \theta ={\sin }^{-1}\left(\frac{x}{a}\right)$

$\frac{d}{dx}\left({\tan }^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)\right)$

$$\begin{gathered} =\frac{d}{dx}\left({\tan }^{-1}\left(\frac{a\sin \left(\theta \right)}{\sqrt{a^2-{\left(a\sin \left(\theta \right)\right)}^2}}\right)\right) \\ =\frac{d}{dx}\left({\tan }^{-1}\left(\frac{a\sin \left(\theta \right)}{a\cos \left(\theta \right)}\right)\right) \\ =\frac{d}{dx}\left({\tan }^{-1}\left(\tan \left(\theta \right)\right)\right) \\ =\frac{d}{dx}\left(\theta \right) \\ =\frac{d}{dx}\left({\sin }^{-1}\left(\frac{x}{a}\right)\right) \end{gathered}$$

Step-2: Differentiate w.r.t $x$

$$\begin{gathered} =\frac{d}{dx}\left({\sin }^{-1}\left(\frac{x}{a}\right)\right) \\ =\frac{1}{\sqrt{1-{\displaystyle \frac{x^2}{a^2}}}}\times \frac{1}{a} \\ =\frac{a}{\sqrt{a^2-x^2}}\times \frac{1}{a} \\ =\frac{1}{\sqrt{a^2-x^2}} \end{gathered}$$

Hence, the correct answer is option $\left(B\right)$.