Question

If $y={\tan }^{-1}\frac{\sqrt{1+x^4-1}}{x},$ then $\frac{dy}{dx}=$

• A. $\frac{1}{1+x^2}$
• B. $\frac{-1}{1+x^2}$
• C. $\frac{1}{2\left(1+x^2\right)}$
• D. $\frac{2}{1+x^2}$

Answer 1

The correct option is A $\frac{1}{1+x^2}$

$y={\tan }^{-1}\frac{\sqrt{1+x^4-1}}{x}$

$\Rightarrow y={\tan }^{-1}\frac{\sqrt{x^4}}{x}$

$\Rightarrow y={\tan }^{-1}\frac{x^2}{x}$

$\Rightarrow y={\tan }^{-1}x$

$\Rightarrow \frac{dy}{dx}=\frac{1}{1+x^2}$