Question

$\frac{d}{dx}{\csc }^{-1}\left(\frac{1+x^2}{2x}\right)$ is equal to

• A. $\frac{-2}{(1+x^2)},x\ne 0$
• B. $\frac{2}{(1+x^2)},x\ne 0$
• C. $\frac{2(1-x^2)}{(1+x^2)|1-x^2|},x\ne \pm 1,0$
• D. None of the above

Answer 1

Answer: (C)

$\frac{2(1-x^2)}{(1+x^2)|1-x^2|},x\ne \pm 1,0$

$\frac{d}{dx}{\csc }^{-1}\left(\frac{1+x^2}{2x}\right)=\frac{d}{dx}{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$

$=\frac{1}{\sqrt{1-{\left(\frac{2x}{1+x^2}\right)}^2}}\left(\frac{(1+x^2)2-\left(2x\right)\left(2x\right)}{{(1+x^2)}^2}\right)$

$=\frac{2(1-x^2)}{(1+x^2)|1-x^2|},$ $x\ne \pm 1,0$