Question

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

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

Answer 1

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

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

$\Rightarrow \left(\frac{1}{{\mu }^2+1}\right)\left(\frac{\sqrt{a^2-x^2}\frac{d}{dx}\left(x\right)\frac{d}{dx}\left(\sqrt{a^2-x^2}\right).x}{{\left(\sqrt{a^2-x^2}\right)}^{}}\right)$

$\Rightarrow \left(\frac{1}{{\mu }^2+1}\right)\left[\frac{\left(\sqrt{a^2-x^2}\right)\left(1\right)-\frac{(-2x)}{2\sqrt{a^2-x^2}}.x}{(a^2-x^2)}\right]$

$\Rightarrow \left(\frac{1}{{\mu }^2+1}\right)(\frac{a^2-x^2}{(a^2-x^2)}+\frac{x^2}{\sqrt{a^2-x^2}})$

$\Rightarrow \frac{1}{[{\left(\frac{x}{\sqrt{a^2-x^2}}\right)}^2+1]}\left(\frac{a^2}{{(a^2-x^2)}^{3/2}}\right)$

$\Rightarrow \frac{(a^2-x^2)}{x^2+(a^2-x^2)}.\frac{a^2}{(a^2-x^2)\left(\sqrt{a^2-x^2}\right)}$

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

Option $D$ is correct.