Question

Let $f\left(x\right)={\cot }^{-1}\left(\frac{1-x^2}{2x}\right)+{\cot }^{-1}\left(\frac{1-3x^2}{3x-x^3}\right)-{\cot }^{-1}\left(\frac{1-6x^2+x^4}{4x-4x^3}\right)$, the $F^{\prime }\left(x\right)$ equals

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

Answer 1

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

$F\left(x\right)={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+{\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)-{\tan }^{-1}\left(\frac{4x-4x^3}{1-6x^2+x^4}\right)$
$=2{\tan }^{-1}x+3{\tan }^{-1}x-4{\tan }^{-1}x={\tan }^{-1}x$
$\Rightarrow F^{\prime }\left(x\right)=\frac{1}{1+x^2}$