Question

$\frac{d}{dx}\left\{co{s}^2\right[Ta{n}^{-1}\left(sinco{t}^{-1}x\right)\left]\right\}$=

• A. $\frac{2}{(x^2+2{)}^2}$
• B. $\frac{2x}{(x^2+2{)}^2}$
• C. $\frac{x^2+1}{x^2+2}$
• D. None

Answer 1

The correct option is B $\frac{2x}{(x^2+2{)}^2}$


$y=co{s}^2\left[ta{n}^{-1}\right(sin\left(co{t}^{-1}x\right)\left)\right]co{t}^{-1}x=zx=cotzsinz=\frac{1}{\sqrt{1+x^2}}z=si{n}^{-1}\frac{1}{\sqrt{1+x^2}}$
$z=si{n}^{-1}\frac{1}{\sqrt{1+x^2}}y=co{s}^2\left[ta{n}^{-1}\right(sin\left(si{n}^{-1}\frac{1}{\sqrt{1+x^2}}\right)\left)\right]$
$=co{s}^2\left[ta{n}^{-1}\frac{1}{\sqrt{1+x^2}}\right]Letta{n}^{-1}\frac{1}{\sqrt{1+x^2}}=utanu=\frac{1}{\sqrt{1+x^2}}y=co{s}^2\left(co{s}^{-1}\frac{\sqrt{1+x^2}}{2+x^2}\right)y=co{s}^2\left(cosco{s}^{-1}\frac{\sqrt{1+x^2}}{2+x^2}\right)\left(cosco{s}^{-1}\frac{\sqrt{1+x^2}}{2+x^2}\right)y=\frac{1+x^2}{2+x^2}\frac{dy}{dx}=\frac{(2+x^2)\left(2x\right)-(1+x^2)\left(2x\right)}{(2+x^2{)}^2}=\frac{4x+2x^3-2x-2x^3}{(2+x^2{)}^2}=\frac{2x}{(2+x^2{)}^2}$.