Question

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

• A. $\frac{1}{\sqrt{a^2+x^2}}$
• B. $\frac{-1}{\sqrt{a^2+x^2}}$
• C. $\frac{x}{\sqrt{a^2-x^2}}$
• D. $\frac{1}{\sqrt{a^2-x^2}}$

Answer 1

The correct option is D $\frac{1}{\sqrt{a^2-x^2}}$

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

from triangle

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

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

Let $y=\frac{{\sin }^{-1}x}{a}$

$\sin y=\frac{x}{a}$ $\cos y=\frac{\sqrt{a^2-x^2}}{a}$

$\cos y\frac{dy}{dx}=\frac{1}{a}$

$\frac{dy}{dx}=\frac{1}{a}\times \frac{a}{\sqrt{a^2-x^2}}=\frac{1}{\sqrt{a^2-x^2}}$

$D$ is correct