The derivative of e^sinx is equal to cosx e^{sinx}. In this post, we will find the derivative of e^{sinx} by first principle.

The e^{sinx} derivative formula is given as follows:

$\dfrac{d}{dx}$(e^{sinx}) = cosx e^{sinx}.

## Derivative of e^sinx Using First Principle

The derivative of f(x) using the first principle is given by the following limit formula:

$\dfrac{d}{dx}$ ( f(x) ) = lim_{h→0} $\dfrac{f(x+h)-f(x)}{h}$.

Put f(x) = e^{sinx}.

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

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

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

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

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

= e^{sinx} lim_{z→0} $\dfrac{e^z-1}{z}$ × lim_{h→0} $\dfrac{2\cos(x+\frac{h}{2})\sin \frac{h}{2}}{h}$, using the formula sina – sinb = 2 cos(a+b)/2 sin(a-b)/2.

= e^{sinx} × 1 × lim_{h→0} $\cos(x+\frac{h}{2})$ × lim_{h→0} $\dfrac{\sin \frac{h}{2}}{\frac{h}{2}}$ as the limit lim_{x→0}(e^{x}-1)/x =1.

= e^{sinx} cos(x+\frac{0}{2}) × lim_{u→0} sinu/u where u=h/2.

= e^{sinx} cosx × 1

= cosx e^{sinx}.

So the derivative of e^sinx is equal to cosx e^{sinx}, and this is obtained by the first principle of derivatives.

**More Derivatives:**

## FAQs

### Q1: What is the derivative of e^sinx?

**Answer:** The derivative of e^sinx (e to the power sinx) is equal to cosx e^{sinx}.