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 e^{sin x}.

**What is the derivative of e**^{sin x}?

^{sin x}?

Let us find the derivative of e^{sin 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 e^{sin x} by chain rule is cos x e^{sin x}. In other words, d/dx(e^{sin x})=cos x e^{sin x}.

**Also Read:**

Integration of mod x : The integration of |x| is -x|x|/2+cDerivative of log : The derivative of log_{e} 3x_{e} 3x is 1/x.Derivative of sin 3x : The derivative of sin 3x is 3cos 3x. |

**Derivative of e**^{sin x} from first principle

^{sin x}from first principle

Now we will find the derivative of e^{sin x} using the limit definition of derivatives. Let f(x)=e^{sin 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 the derivative of e^{sinx} (e to the power sin x) from first principle is equal to e^{sinx} cosx.

**Also Read:**

Derivative of e: The derivative of e^{cos x}^{cos x }is -sin x e^{cos x}.Derivative of square root of x: The derivative of root x is 1/2√x. |

**Derivative of e**^{sin x} by substitution method

^{sin 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=e^{sin x}. We need to find du/dx.

Step 2: Taking logarithm on both sides, we get

log_{e}u = log_{e}e^{sin x}

⇒ log_{e}u = sin x log_{e}e

⇒ log_{e}u = sin x as we know that log_{a}a=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=e^{sin x}

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

**Application of the derivative of e**^{sin x}

^{sin x}

As an application, we will find the differentiation of e^{sin 2x}. Let f(x)=e^{sin x}. So we have f(2x)=e^{sin 2x}.

From above we have: $\dfrac{d}{dx}\left( f(x) \right)$=cos x e^{sin 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 e^{sin 2x} is 2cos 2x e^{sin 2x}.

**Also Read:**

Derivative of xlogx: The derivative of xlogx^{ }is 1+logx.Derivative of xe: The derivative of root xe^{x}^{x} is (1+x)e^{x}. |

**FAQs on Derivative of e**^{sin x}

^{sin x}

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

Ans: Applying the chain rule of derivatives, we get that d/dx(e^{sin x}) = e^{sin x} d/dx(sin x) = cos x e^{sin 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 e**

^{x}sin x?Ans: By the product rule of derivatives, we have d/dx(e^{x}sin x)= d/dx(e^{x}) sin x + e^{x} d/dx(sin x) = e^{x}sin x+e^{x}cos x. Thus the derivative of e^{x}sin x is equal to the product of e^{x} and sin x+cos x.