# Derivative of e^sinx: Formula, Proof by First Principle, Chain Rule

The derivative of esinx is equal to cosx esinx. Here we will calculate the derivative of e to the power sin x by the first principle, using the substitution method and the chain rule of derivatives. As an application, we will find the differentiation of several functions involving esin x.

## What is the derivative of esin x?

Let us find the derivative of esin x by the chain rule of derivatives. Recall the chain rule: $\dfrac{df}{dx}=\dfrac{df}{du} \cdot \dfrac{du}{dx}$, where f is a function of u.

Step 1: Put u=sin x

Step 2: Differentiating with respect to x, we get

$\dfrac{du}{dx}$ = cos x

Step 3: Now, $\dfrac{d}{dx}\left(e^{\sin x} \right)=\dfrac{d}{dx}\left(e^u \right)$

$=\dfrac{d}{du}\left(e^u \right) \cdot \dfrac{du}{dx} \quad$ by the chain rule

= eu cosx

= cosx esinx as u=sin x.

So the derivative of esin x by chain rule is cos x esin x. In other words, d/dx(esin x)=cos x esin x.

## Derivative of esin x from first principle

Now we will find the derivative of esin x using the limit definition of derivatives. Let f(x)=esin x. Then the derivative of f(x) by first principle is given as follows:

$\dfrac{d}{dx}\left( f(x) \right)$ = limh→0 $\dfrac{f(x+h)-f(x)}{h}$

So $\dfrac{d}{dx}\left( e^{\sin x} \right)$ = limh→0 $\dfrac{e^{\sin(x+h)} – e^{\sin x}}{h}$

= limh→0 $\dfrac{e^{\sin x}(e^{\sin(x+h)-\sin x} -1)}{h}$

= esinx limh→0 $\dfrac{e^{\sin(x+h)-\sin x} -1}{h}$

= esinx limh→0 $\dfrac{e^{\sin(x+h)-\sin x} -1}{\sin(x+h)-\sin x}$ × limh→0 $\dfrac{\sin(x+h)-\sin x}{h}$

[Let t=sin(x+h)-sin x. Then t→0 as h→0]

= esinx limt→0 $\dfrac{e^t-1}{t}$ × limh→0 $\dfrac{2\cos(x+h/2)\sin h/2}{h}$ as we know that sin c – sin d= 2 cos(c+d)/2 sin(c-d)/2.

= esinx × 1 × limh→0 $\cos(x+h/2)$ × limh→0 $\dfrac{\sin h/2}{h/2}$

= esinx cos(x+0/2) × limz→0 sinz/z where z=h/2.

= esinx cosx × 1 as limx→0 sinx/x=1

= cosx esinx

So the derivative of esinx (e to the power sin x) from first principle is equal to esinx cosx.

## Derivative of esin x by substitution method

Next, we evaluate the derivative of e to the power sin x by the substitution method. Here we will use the logarithmic derivatives.

Step 1: Let u=esin x. We need to find du/dx.

Step 2: Taking logarithm on both sides, we get

logeu = logeesin x

⇒ logeu = sin x logee

⇒ logeu = sin x as we know that logaa=1.

Step 3: Differentiating we get that

$\dfrac{d}{dx}\left(\log_e u \right)=\dfrac{d}{dx}\left(\sin x \right)$

⇒ $\dfrac{1}{u} \dfrac{du}{dx}$ = cos x

⇒ $\dfrac{du}{dx}$ = u cosx = esinx cosx as u=esin x

Thus, we obtain the derivative of e power sin x by the logarithmic differentiation which is cos x esin x.

## Application of the derivative of esin x

As an application, we will find the differentiation of esin 2x. Let f(x)=esin x. So we have f(2x)=esin 2x.

From above we have: $\dfrac{d}{dx}\left( f(x) \right)$ = cos x esin x.

Note that we have to find the derivative of f(2x). So we put z=2x and hence dz/dx=2.

Now the derivative of f(2x) is equal to

$\dfrac{d}{dx}\left( f(2x) \right)=\dfrac{d}{dx}\left( f(z) \right)$

$=\dfrac{d}{dz}\left( f(z) \right) \cdot \dfrac{dz}{dx} \quad$ by the chain rule

= cosz esinz × 2

= 2cos2x esin2x as z=2x.

Hence the derivative of esin2x is 2cos2x esin2x.

## FAQs on Derivative of esin x

Q1. Find the derivative of e to the power sin x.

Ans: Applying the chain rule of derivatives, we get that d/dx(esin x) = esin x d/dx(sin x) = cos x esin x. So the derivative of e to the power sin x is equal to the product of cos x and e to the power sin x

Q2. What is the derivative of exsin x?

Ans: By the product rule of derivatives, we have d/dx(exsin x)= d/dx(ex) sin x + ex d/dx(sin x) = exsin x+excos x. Thus the derivative of exsin x is equal to the product of ex and sin x+cos x.

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