Question

differentiate e^2cosx​

Answer 1

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

Given function is

$\rm :\longmapsto\: {e}^{2cosx}$

Let assume that

$\rm :\longmapsto\: y = {e}^{2cosx}$

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

$\rm :\longmapsto\: \dfrac{d}{dx}y = \dfrac{d}{dx}{e}^{2cosx}$

We know,

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

So, using this, we get

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

We know,

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

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {e}^{2cosx} \times 2\dfrac{d}{dx}(cosx)$

We know,

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

So, using this, we get

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

$\bf\implies \:\boxed{ \bf{ \: \dfrac{dy}{dx} = - \: 2 \: sinx \: {e}^{2cosx}}}$

More to Know :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Answer:

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

Given function is

$\rm :\longmapsto\: {e}^{2cosx}$

Let assume that

$\rm :\longmapsto\: y = {e}^{2cosx}$

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

$\rm :\longmapsto\: \dfrac{d}{dx}y = \dfrac{d}{dx}{e}^{2cosx}$

We know,

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

So, using this, we get

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

We know,

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

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {e}^{2cosx} \times 2\dfrac{d}{dx}(cosx)$

We know,

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

So, using this, we get

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

$\bf\implies \:\boxed{ \bf{ \: \dfrac{dy}{dx} = - \: 2 \: sinx \: {e}^{2cosx}}}$

More to Know :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$