Question

If $y={\tan }^{-1}\left(\frac{2x-1}{1+x-x^2}\right)$, then $\frac{dy}{dx}$ at $x=1$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $1$
• D. $\frac{-1}{2}$
• E. $\frac{3}{2}$

Answer 1

The correct option is E $\frac{3}{2}$

Given,
$y={\tan }^{-1}\left(\frac{2x-1}{1+x-x^2}\right)={\tan }^{-1}\left\{\frac{x-(1-x)}{1+x(1-x)}\right\}$
$\Rightarrow y={\tan }^{-1}x-{\tan }^{-1}(1-x)$
On differentiating w.r.t. $x$, we get
$\frac{dy}{dx}=\frac{1}{1+x^2}-\frac{1}{1+(1-x{)}^2}(-1)$
$=\frac{1}{1+x^2}+\frac{1}{1+(1-x{)}^2}$
At $x=1,{\left[\frac{dy}{dx}\right]}_{x=1}=\frac{1}{1+1}+\frac{1}{1+0}=\frac{1}{2}+1=\frac{3}{2}$.