Derivative of e^sin x

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.

Also Read:

Integration of mod x : The integration of |x| is -x|x|/2+c

Derivative of loge 3x : The derivative of loge 3x is 1/x.

Derivative of sin 3x : The derivative of sin 3x is 3cos 3x.

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.

Also Read:

Derivative of ecos x: The derivative of ecos x is -sin x ecos x.

Derivative of square root of x: The derivative of root x is 1/2√x.

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.

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.