# Derivative of e^sin x | e^sinx Derivative

Here we will calculate the derivative of e to the power sin x by the method of 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

$=e^u \cdot \cos x$

$=\cos x e^{\sin x}$ as u=sin x.

So the derivative of esin x 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:

$\dfrac{d}{dx}\left( f(x) \right)$ $= \lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

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

$=\lim\limits_{h \to 0} \dfrac{e^{\sin x}(e^{\sin(x+h)-\sin x} -1)}{h}$

$=e^{\sin x}\lim\limits_{h \to 0} \dfrac{e^{\sin(x+h)-\sin x} -1}{h}$

$=e^{\sin x}\lim\limits_{h \to 0} \dfrac{e^{\sin(x+h)-\sin x} -1}{\sin(x+h)-\sin x}$ $\times \lim\limits_{h \to 0} \dfrac{\sin(x+h)-\sin x}{h}$

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

$=e^{\sin x} \lim\limits_{t \to 0} \dfrac{e^t-1}{t}$ $\times \lim\limits_{h \to 0} \dfrac{2\cos(x+h/2)\sin h/2}{h}$ as we know that sin c – sin d= 2 cos(c+d)/2 sin(c-d)/2.

$=e^{\sin x} \times 1$ $\times \lim\limits_{h \to 0}\cos(x+h/2)$ $\times \lim\limits_{h \to 0} \dfrac{\sin h/2}{h/2}$

$=e^{\sin x} \cos(x+0/2)$ $\times \lim\limits_{z \to 0} \dfrac{\sin z}{z}$ where z=h/2.

$=e^{\sin x} \cos x \times 1$ as $\lim\limits_{x \to 0} \dfrac{\sin x}{x}=1$

$=\cos x e^{\sin x}$

So we have obtained the derivative of e to the power sin x from first principle.

## 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

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

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

$\Rightarrow \dfrac{du}{dx}=u \cos x =e^{\sin x} \cos x$ as u=esin x

Thus, we obtain the derivative of e power sin x 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

$=\cos z e^{\sin z} \cdot 2$

$=2\cos 2x e^{\sin 2x}$ as z=2x.

Hence the derivative of esin 2x is 2cos 2x esin 2x.