Question

given that y=e^x /e^x-1 find dy/dx

Answer 1

Answer:

based on dy/dx e^x=e^x is used

Answer 2

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

➢ Given that

$\rm :\longmapsto\:y = \dfrac{ {e}^{x} }{{e}^{x} - 1}$

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

$\rm :\longmapsto\:\dfrac{d}{dx} y =\dfrac{d}{dx} \: \dfrac{ {e}^{x} }{{e}^{x} - 1}$

We know that

$\underbrace{\boxed{\sf{\dfrac{d}{dx} \frac{u}{v} \: = \: \frac{v\dfrac{d}{dx}u \: - \: u\dfrac{d}{dx}v}{ {v}^{2} } }}}$

So, using this formula, we have

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{({e}^{x} - 1)\dfrac{d}{dx}{e}^{x} \: - \: {e}^{x}\dfrac{d}{dx}({e}^{x} - 1)}{ {({e}^{x} - 1)}^{2} }$

We know that,

$\underbrace{\boxed{\sf{\dfrac{d}{dx}{e}^{x} = {e}^{x}}}} \: \: \: \: \: \rm and \: \: \: \: \: \underbrace{\boxed{\sf{ \dfrac{d}{dx}k = 0}}}$

So, using these, we have

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{({e}^{x} - 1){e}^{x} \: - \: {e}^{x}({e}^{x} - 0)}{ {({e}^{x} - 1)}^{2} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{{e}^{2x} - {e}^{x} \: - \: {e}^{2x}}{ {({e}^{x} - 1)}^{2} }$

$\bf :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ - {e}^{x}}{ {({e}^{x} - 1)}^{2} }$

Second Method :-

➢ Given that

$\rm :\longmapsto\:y = \dfrac{ {e}^{x} }{{e}^{x} - 1}$

can be rewritten as

$\rm :\longmapsto\:y = \dfrac{ {e}^{x} - 1 + 1 }{{e}^{x} - 1}$

$\rm :\longmapsto\:y = \dfrac{ {e}^{x} - 1 }{{e}^{x} - 1} + \dfrac{1}{{e}^{x} - 1}$

$\rm :\longmapsto\:y = 1 + {({e}^{x} - 1)}^{ - 1}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx}( 1 + {({e}^{x} - 1)}^{ - 1} )$

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{d}{dx}1 + \dfrac{d}{dx}{({e}^{x} - 1)}^{ - 1}$

$\underbrace{\boxed{\sf{\dfrac{d}{dx}{x}^{n} = {nx}^{n - 1}}}} \: \: \: \: \: \rm and \: \: \: \: \: \underbrace{\boxed{\sf{ \dfrac{d}{dx}k = 0}}}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = 0 + ( - 1) {({e}^{x} - 1)}^{ - 1 - 1}\dfrac{d}{dx}({e}^{x} - 1)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = - 1 {({e}^{x} - 1)}^{ -2}{e}^{x}$

$\bf :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ - {e}^{x}}{ {({e}^{x} - 1)}^{2} }$