Question

Let $f\left(x\right)={\sin }^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)$, find the value of $\frac{d\left(\tan f\right(x\left)\right)}{d\left({\tan }^{-1}x\right)}$, when $x=\frac{1}{\sqrt{2}}$

Answer 1

Let, $f\left(x\right)={\sin }^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)$
$={\tan }^{-1}\left(\frac{2x+2}{\sqrt{(4x^2+8x+13)-(2x+3{)}^2}}\right)$
$={\tan }^{-1}\left(\frac{2x+2}{3}\right)$
$\therefore \frac{d\left(\tan f\right(x\left)\right)}{d\left({\tan }^{-1}x\right)}=\frac{d\left(\frac{2x+2}{3}\right)}{d\left({\tan }^{-1}x\right)}$
$=\frac{\frac{2}{3}}{\frac{1}{1+x^2}}$.
Now, $\frac{d\left(\tan f\right(x\left)\right)}{d\left({\tan }^{-1}x\right)}$ at $x=\frac{1}{\sqrt{2}}$ $=\frac{\frac{2}{3}}{\frac{1}{1+\frac{1}{2}}}=\frac{\frac{2}{3}}{\frac{2}{3}}=1$.