Question

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

• A. $2x[1+tan(logx)]+xse{c}^2(logx)$
• B. $x[1+tan(logx)]+se{c}^2(logx)$
• C. $2x[1+tan(logx)]+x^{2}se{c}^2(logx)$
• D. $2x[1+tan(logx)]+se{c}^2(logx)$

Answer 1

Answer: (A)

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