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

The derivative formula of e^{cosx} is given below:

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

## Derivative of e^cosx Using First Principle

By first principle, the derivative of a function 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^{cosx}.

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

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

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

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

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

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

= e^{cosx} × 1 × lim_{h→0} $- \sin(x+\frac{h}{2})$ × lim_{h→0} $\dfrac{\sin \frac{h}{2}}{\frac{h}{2}}$ using the limit formula lim_{t→0}(e^{t}-1)/t =1.

= – e^{cosx} $\sin(x+\frac{0}{2})$ × lim_{u→0} sinu/u where u=h/2.

= – sinx e^{cosx} as the limit of sinx/x is 1 when x→0.

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

**More Derivatives:** Derivative of e^{sinx} by first principle

## FAQs

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

**Answer:** The derivative of e^cosx is equal to -sinx e^{cosx}.