# Find the derivative

1. May 10, 2010

### npellegrino

I am having difficulty figuring this one, any guidance will be appreciated.

2. May 10, 2010

### Cyosis

We cannot guide you until you show us where you get stuck. What kind of methods have you tried to use so far?

3. May 10, 2010

### npellegrino

This is what I have so far.

I believe i use this rule d/dx log a ^ u = 1/(lna)u

so ln = 1
a = ln
u = (1+e^sqrtX)/(2-e^cosx)

now i need to take the derivative of u, does e^sqrtX = e^sqrtX or is it e^sqrtX * derivative of sqrtX making it e^sqrtX * 1/2sqrtX

d/dx u =

4. May 10, 2010

### Cyosis

This 'rule' makes no sense whatsoever.

This is also pretty hard to read. Yes $\exp(\sqrt{x})=\exp(\sqrt{x})$. This is no surprise since everything equals itself. What you probably mean is $\frac{d}{dx}\exp(\sqrt{x})=\exp(\sqrt{x})$. That is wrong since you have to use the chain rule therefore your second guess,$\frac{d}{dx}\exp(\sqrt{x})=\exp(\sqrt{x}) \frac{1}{2 \sqrt{x}}$, is correct.