Question

Differentiate the above with clear steps.

Answer 1

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

Given function is

$\rm :\longmapsto\: {\bigg(sin }^{\dfrac{5}{2} }{e}^{x}\bigg)$

Let assume that,

$\rm :\longmapsto\:y = {\bigg(sin }^{\dfrac{5}{2} }{e}^{x}\bigg)$

can be rewritten as

$\rm :\longmapsto\:y = {\bigg(sin{e}^{x}\bigg) }^{\dfrac{5}{2} }$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx} {\bigg(sin{e}^{x}\bigg) }^{\dfrac{5}{2} }$

We know,

$\boxed{ \bf{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}$

Using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{5}{2} {\bigg(sin{e}^{x}\bigg) }^{\dfrac{5}{2} - 1} \: \dfrac{d}{dx}sin{e}^{x}$

We know,

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

So using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{5}{2} {\bigg(sin{e}^{x}\bigg) }^{\dfrac{3}{2}} \: cos{e}^{x}\dfrac{d}{dx}{e}^{x}$

We know,

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

So, using this we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{5}{2} {\bigg(sin{e}^{x}\bigg) }^{\dfrac{3}{2}} \: cos{e}^{x} \times {e}^{x}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{5 \: {e}^{x} \: cos{e}^{x}}{2} {\bigg(sin{e}^{x}\bigg) }^{\dfrac{3}{2}} \:$

Additional Information :-

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

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

.

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

$\green{\boxed{ \bf{ \dfrac{d}{dx} log_{a}(x) = \frac{1}{x \: loga}}}}$

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

$\green{\boxed{ \bf{ \dfrac{d}{dx}k \: f(x) = k \: \dfrac{d}{dx}f(x)}}}$

$\green{\boxed{ \bf{ \dfrac{d}{dx}u.v \: = \: u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u}}}$