# Derivatives cont'd...

• Feb 24th 2010, 02:47 PM
Jools
Derivatives cont'd...
I have another derivative question, this one's a little more complicated...

Find the derivative of the following:

h(x) = 3e^((sin)(x+2))

I'm having a hard time even getting started on this one...
• Feb 24th 2010, 04:18 PM
satx
Quote:

Originally Posted by Jools
I have another derivative question, this one's a little more complicated...

Find the derivative of the following:

h(x) = 3e^((sin)(x+2))

I'm having a hard time even getting started on this one...

This is precalc? Wow. We didn't even touch derivatives when I took it. Good luck...
• Feb 25th 2010, 12:25 AM
mathemagister
Quote:

Originally Posted by Jools
I have another derivative question, this one's a little more complicated...

Find the derivative of the following:

h(x) = 3e^((sin)(x+2))

I'm having a hard time even getting started on this one...

What do you mean by (sin)(x+2)? Do you mean:

$\displaystyle h(x) = 3e^{\sin(x+2)}$

Using the chain rule: the derivative of $\displaystyle e^{f(x)}$ is $\displaystyle e^{f(x)} \cdot f'(x)$

$\displaystyle h'(x) = 3e^{\sin(x+2)} \cdot \cos(x+2) \cdot 1 = 3 \cos(x+2) e^{\sin(x+2)}$
• Feb 25th 2010, 12:29 AM
mathemagister
Quote:

Originally Posted by Jools
I have another derivative question, this one's a little more complicated...

Find the derivative of the following:

h(x) = 3e^((sin)(x+2))

I'm having a hard time even getting started on this one...

In case you meant:

$\displaystyle h(x) = 3e^{(\sin{x})(x+2)}$

Then, using the same method I did before and the product rule:

$\displaystyle h'(x) = 3e^{(\sin{x})(x+2)} [(\sin{x}) + (x+2)\cos{x}]$

Do you understand it now? :)
• Feb 25th 2010, 06:05 AM
Jools
Thank you! I do... There's just so many varaitions of the chain rule, I have a hard time determining which one to use... I guess practise will be the key.
• Feb 26th 2010, 04:05 AM
HallsofIvy
There is only one chain rule- applied to many different functions.