Question

find derivative of y=log e^x/x

Answer 1

Step-by-step explanation:

here the answer any doubt ask

Answer 2

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

➢ Given that

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

We know that

$\underbrace{\boxed{\sf{log{e}^{x} = x}}}$

So, using this we get

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

$\rm :\longmapsto\:y = 1$

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

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

We know that

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

So, using this

$\bf :\longmapsto\:\dfrac{dy}{dx} = 0$

Hence,

$\bf :\longmapsto\:\dfrac{d}{dx} \: \dfrac{log{e}^{x}}{x} = 0$

Additional Information :-

Let take one more example!!

Question :- Differentiate the following :-

$\rm :\longmapsto\:y = {e}^{ log(x + \sqrt{ {x}^{2} + 1} )}$

Solution :-

$\rm :\longmapsto\:y = {e}^{ log(x + \sqrt{ {x}^{2} + 1} )}$

We know that

$\underbrace{\boxed{\sf{{e}^{logx} = x}}}$

So, using this

$\rm :\longmapsto\:y = x + \sqrt{ {x}^{2} + 1}$

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

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

$\rm :\longmapsto\:\dfrac{dy}{dx}= \dfrac{d}{dx}x + \dfrac{d}{dx}\sqrt{ {x}^{2} + 1}$

We know,

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

and

$\underbrace{\boxed{\bf{ \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} } }}}$

So, using this, we get

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

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

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

$\bf :\longmapsto\:\dfrac{dy}{dx} = 1 + \dfrac{x}{\sqrt{ {x}^{2} + 1 } }$