Question

If derivative of ${\tan }^{-1}\left(\frac{4\sqrt{x}\cdot x^2}{4-x^5}\right)$ is $g\left(x\right)$ for some $x\in R,$ then $g\left(1\right)$ equal to

• A. $1$
• B. $10$
• C. $2$
• D. $8$

Answer 1

The correct option is C $2$

Let $y={\tan }^{-1}\left(\frac{4\sqrt{x}\cdot x^2}{4-x^5}\right)={\tan }^{-1}\left(\frac{\sqrt{x}\cdot x^2}{1-\frac{x^5}{4}}\right)$
$={\tan }^{-1}\left(\frac{2\cdot \sqrt{x}\cdot \frac{x^2}{2}}{1-{(\sqrt{x}\cdot \frac{x^2}{2})}^2}\right)$
$=2{\tan }^{-1}(\sqrt{x}\cdot \frac{x^2}{2})$
$(\because {\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=2{\tan }^{-1}x)$
$=2{\tan }^{-1}\left(\frac{x^{5/2}}{2}\right)$
and $g\left(x\right)=\frac{dy}{dx}$ (given)
$\therefore g\left(x\right)=2\cdot \frac{1}{1+{\left(\frac{x^{5/2}}{2}\right)}^2}\cdot \frac{1}{2}\cdot \frac{5}{2}{x}^{3/2}$
$\Rightarrow g\left(x\right)=\frac{10\cdot x^{3/2}}{4+x^5}$
$\therefore g\left(1\right)=2$