Question

$\sf{if \: x = exp\bigg[tan^{-1}\bigg(\frac{y-x^{2}}{x^{2}}\bigg)\bigg] \: \: Then \: \: \frac{dy}{dX} \: \: equals }$

Options are ::-
$\sf{(a) 2x[1+ tan (log x)]+ x sec²(log x)}$
$\sf{(b) x[1+ tan (log x)]+ sec² (log x)}$
$\sf{(c) 2x[1 + tan (log x)] +x² sec² (log \: x)}$
$\sf{(d) 2x[1+ tan (log \: x)]+ sec²(log \: x )}$

​

Answer 1

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \bf \: \dfrac{d}{dx} tanx \: = {sec}^{2}x}$

$\boxed{\bf \: \dfrac{d}{dx}logx = \dfrac{1}{x}}$

$\boxed{\bf \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}$

$\boxed{\bf \: \dfrac{d}{dx}x = 1}$

$\boxed{\bf \:loge = 1}$

$\boxed{\bf\:\dfrac{d}{dx}u.v = v\dfrac{d}{dx} u \: + \: u\dfrac{d}{dx} v}$

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\sf{\: x = exp\bigg[tan^{-1}\bigg(\frac{y-x^{2}}{x^{2}}\bigg)\bigg]}$

On taking log on both sides, we get

$\rm :\longmapsto\:logx = {tan}^{ - 1}\bigg( \dfrac{y - {x}^{2} }{ {x}^{2} } \bigg) log(e)$

$\rm :\longmapsto\:logx = {tan}^{ - 1}\bigg( \dfrac{y - {x}^{2} }{ {x}^{2} } \bigg) \times 1$

$\rm :\longmapsto\: \tan(logx) = \dfrac{y - {x}^{2} }{ {x}^{2} }$

$\rm :\longmapsto\: {x}^{2} \tan(logx) = y - {x}^{2}$

$\rm :\longmapsto\: {x}^{2} \tan(logx) + {x}^{2} = y$

$\rm :\longmapsto\: {x}^{2}( \tan(logx) + 1) = y$

$\rm :\longmapsto\:y = {x}^{2}( \tan(logx) + 1)$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\: \dfrac{d}{dx} y = \dfrac{d}{dx} ( {x}^{2}( \tan(logx) + 1))$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {x}^{2}\dfrac{d}{dx} ( \tan(logx) + 1) + (\tan(logx) + 1)\dfrac{d}{dx} {x}^{2}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {x}^{2} {sec}^{2}(logx)\dfrac{d}{dx}logx + ( \tan(logx) + 1)2x$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {x}^{2} {sec}^{2}(logx)\dfrac{1}{x} + ( \tan(logx) + 1)2x$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {x}{sec}^{2}(logx) + ( \tan(logx) + 1)2x$

$\bf\implies \:\dfrac{dy}{dx} = 2x \bigg(1 + tan(logx) \bigg) + x {sec}^{2}(logx)$

$\large{\boxed{\boxed{\bf{Option \: (a) \: is \: correct}}}}$

Additional Information :-

$\boxed{\bf\:\dfrac{d}{dx}sinx = cosx}$

$\boxed{\bf\:\dfrac{d}{dx}cosx = - sinx}$

$\boxed{\bf\:\dfrac{d}{dx}cotx = - {cosec}^{2}x}$

$\boxed{\bf\:\dfrac{d}{dx}secx = secxtanx}$

$\boxed{\bf\:\dfrac{d}{dx}cosecx = - cosecxcotx}$