Question

If x=exp\left\\{tan^{-1}\left(\frac{y-x^{2}}{x^{2}}\right)\right\\}, then $\frac{dy}{dx}$ equals

• A. $2x[1+ tan(logx)] + x sec^2(log\,x)$
• B. $x[1 + tan(logx)] + sec^2(log \,x)$
• C. $2x[1 + tan(logx)] + x^2sec^2(log\,x)$
• D. $2x[1 + tan(logx)] + sec^2(log\,x)$

Answer 1

Answer: (A)

$2x[1+ tan(logx)] + x sec^2(log\,x)$

Answer 2

Given that, x=exp\left\\{tan^{-1}\left(\frac{y-x^{2}}{x^{2}}\right)\right\\} Taking log on both sides, we get $log\,x=tan^{-1}\left(\frac{y-x^{2}}{x^{2}}\right)$ $\Rightarrow \frac{y-x^{2}}{x^{2}}=tan\left(log\,x\right)$ $\Rightarrow y=x^{2}\,tan\left(logx\right)+x^{2}$ Differentiating $w$.$r$.$t$. $x$, we get $\frac{dy}{dx}=2x\,tan\left(log\,x\right)+x^{2} \frac{sec^{2}\left(log\,x\right)}{x}+2x$ $\Rightarrow \frac{dy}{dx}=2x\left[1+tan\left(log\,x\right)\right]+x\,sec^{2}\left(log\,x\right)$