Question

differentiate wrt x, find dy/dx.​

Answer 1

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

Given that

$\rm :\longmapsto\:y = \sqrt[3]{\dfrac{4x - 1}{(2x + 3) {(5 - 2x)}^{2} } }$

Taking log on both sides, we get

$\rm :\longmapsto\:logy = log \bigg( \sqrt[3]{\dfrac{4x - 1}{(2x + 3) {(5 - 2x)}^{2} } } \bigg)$

We know,

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

Using this, we get

$\rm :\longmapsto\:logy = \dfrac{1}{3} log \bigg(\dfrac{4x - 1}{(2x + 3) {(5 - 2x)}^{2} } \bigg)$

We know,

$\red{\rm :\longmapsto\: log(xy) = log(x) + log(y) }$

$\red{\rm :\longmapsto\: log(\dfrac{x}{y} ) = log(x) - log(y) }$

So, using this, we get

$\rm :\longmapsto\:logy = \dfrac{1}{3}\bigg(log(4x - 1) - log(2x + 3) - 2log(5 - 2x)\bigg)$

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

$\rm :\longmapsto\:\dfrac{d}{dx} logy =\dfrac{d}{dx} \dfrac{1}{3}\bigg(log(4x - 1) - log(2x + 3) - 2log(5 - 2x)\bigg)$

We know,

$\red{\rm :\longmapsto\:\dfrac{d}{dx}k \: f(x) = k\dfrac{d}{dx}f(x)}$

So using this,

$\rm :\longmapsto\:\dfrac{1}{y} \dfrac{dy}{dx} =\dfrac{1}{3} \dfrac{d}{dx}\bigg(log(4x - 1) - log(2x + 3) - 2log(5 - 2x)\bigg)$

We know,

$\boxed{ \rm{ \dfrac{d}{dx}log(ax + b) = \frac{a}{ax + b}}}$

So, using this,

$\rm :\longmapsto\:\dfrac{1}{y} \dfrac{dy}{dx} =\dfrac{1}{3} \bigg(\dfrac{4}{4x - 1} - \dfrac{2}{2x + 3} + \dfrac{4}{5 - 2x} \bigg)$

$\rm :\longmapsto\: \dfrac{dy}{dx} =\dfrac{y}{3} \bigg(\dfrac{4}{4x - 1} - \dfrac{2}{2x + 3} + \dfrac{4}{5 - 2x} \bigg)$

Additional Information :-

$\boxed{ \rm{ \dfrac{d}{dx}(ax + b) = a}}$

$\boxed{ \rm{ \dfrac{d}{dx}sin(ax + b) = acos(ax + b)}}$

$\boxed{ \rm{ \dfrac{d}{dx}cos(ax + b) = - asin(ax + b)}}$

$\boxed{ \rm{ \dfrac{d}{dx}cosec(ax + b) = - acosec(ax + b) \: cot(ax + b)}}$

$\boxed{ \rm{ \dfrac{d}{dx}sec(ax + b) = a \: sec(ax + b) \: tan(ax + b)}}$

$\boxed{ \rm{ \dfrac{d}{dx}tan(ax + b) = a \: sec^{2} (ax + b) \: }}$

$\boxed{ \rm{ \dfrac{d}{dx}cot(ax + b) = - \: a \: cosec^{2} (ax + b) \: }}$