Question

if y = x^y
then prove that
dy/dx = y² / x(1-y . log x)
using logarithmic differentiation.
​

Answer 1

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

$\boxed{ \red{\bf{\dfrac{d}{dx} log(x) = \dfrac{1}{x} }}}$

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

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

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

Given that

$\rm :\longmapsto\:y = {x}^{y}$

On taking log on both sides, we get

$\rm :\longmapsto\:logy = log \: {x}^{y}$

$\rm :\longmapsto\:logy \: = \: y \: logx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}logy \: = \dfrac{d}{dx}( \: y \: logx \: )$

$\rm :\longmapsto\:\dfrac{1}{y} \dfrac{d}{dx}y = y\dfrac{d}{dx}logx \: + \: logx\dfrac{d}{dx}y$

$\rm :\longmapsto\:\dfrac{1}{y} \dfrac{dy}{dx}= y\dfrac{1}{x} \: + \: logx\dfrac{dy}{dx}$

$\rm :\longmapsto\:\dfrac{1}{y} \dfrac{dy}{dx} \: - \: logx\dfrac{dy}{dx}= \dfrac{y}{x} \:$

$\rm :\longmapsto\: \bigg(\dfrac{1}{y} \: - \: logx \bigg)\dfrac{dy}{dx}= \dfrac{y}{x} \:$

$\rm :\longmapsto\: \bigg(\dfrac{1 - ylogx}{y} \: \bigg)\dfrac{dy}{dx}= \dfrac{y}{x} \:$

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{ {y}^{2} }{x(1 - ylogx)}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

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

$\boxed{ \red{\bf{\dfrac{d}{dx}k = 0}}}$

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

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

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

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

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

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