Question

if y=
x^x^-°°, then
x(1-ylogy) dy is equal to​

Answer 1

Appropriate Question :-

$\bf :\longmapsto\:If \: y = {x}^{ {x}^{ {x - - - \infty }}}, then \: x(1 - ylogx)\dfrac{dy}{dx} =$

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

Given that,

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

can be rewritten as

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

On taking log on both sides,

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

$\rm :\longmapsto\:logy = ylog{x}$

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

$\rm :\longmapsto\:\dfrac{d}{dx} logy = \dfrac{d}{dx} ylog{x}$

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

$\: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \:\dfrac{d}{dx}u.v \: = \: u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u \bigg \}}$

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

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: \dfrac{d}{dx}logx = \dfrac{1}{x} \bigg \}}$

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

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

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

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

Additional Information :-

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

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

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

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

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

$\green{\boxed{ \tt \: \dfrac{d}{dx}secx = \: secx \: tanx}}$

$\green{\boxed{ \tt \: \dfrac{d}{dx}cosecx = - \: cosecx \: cotx}}$

$\green{\boxed{ \tt \: \dfrac{d}{dx} {e}^{x} = {e}^{x} }}$

$\green{\boxed{ \tt \: \dfrac{d}{dx} {a}^{x} = {a}^{x}loga }}$

$\green{\boxed{ \tt \: \dfrac{d}{dx}k = 0}}$