Question

Diffrenciate [tex] {a}^{x} [\tex]
w.r.t. [tex]log_{a}(x)[\tex]

please answer fast
And correct give correct answer please​

Answer 1

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

Let assume that

$\red{\rm :\longmapsto\:u = {a}^{x}}$

and

$\red{\rm :\longmapsto\:v = log_{a}(x)}$

Consider,

$\red{\rm :\longmapsto\:u = {a}^{x}}$

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

${\rm :\longmapsto\:\dfrac{d}{dx} u =\dfrac{d}{dx} {a}^{x}}$

${\rm :\longmapsto\:\dfrac{du}{dx} = {a}^{x}} log(a) - - - (1)$

Consider,

$\red{\rm :\longmapsto\:v = log_{a}(x)}$

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

${\rm :\longmapsto\:\dfrac{d}{dx}v = \dfrac{d}{dx} log_{a}(x)}$

${\rm :\longmapsto\:\dfrac{dv}{dx} = \dfrac{1}{x log(a)}} - - - (2)$

Since, we have to differentiate u w. r. t. v,

So,

$\rm :\longmapsto\:\dfrac{du}{dv} = \dfrac{du}{dx} \div \dfrac{dv}{dx}$

$\rm :\longmapsto\:\dfrac{du}{dv} = {a}^{x} log(a) \div \dfrac{1}{x log(a) }$

$\bf:\longmapsto\:\dfrac{du}{dv} = x \: {a}^{x}{(loga)}^{2}$

Additional Information :-

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

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

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

$\green{\boxed{ \bf{ \:\dfrac{d}{dx} {a}^{x} = {a}^{x} log(a) }}}$

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

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

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

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

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