Question

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

• A. $\frac{-2}{1+x^2}\forall x$
• B. $\frac{-2}{1+x^2}\forall |x|>1$
• C. $\frac{-2}{1+x^2}\forall |x|<1$
• D. none to these

Answer 1

Answer: (C)

$\frac{-2}{1+x^2}\forall |x|<1$

Given

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

$\cos y=\frac{2x}{1+x^2}$

Differentiate on both sides w.r.t $x,$ we get

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

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

$-\sqrt{\frac{(1-x^2{)}^2}{(1+x^2{)}^2}}\frac{dy}{dx}=\frac{2(1-x^2)}{(1+x^2{)}^2}$

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

$\therefore 1-x^2>0,x^2<1$

$\therefore \left|x\right|<1$

$\therefore \frac{dy}{dx}=\frac{-2}{(1+x^2)}$ $\forall \left|x\right|<1$