# First and second derivative of e^6e^x?

• Feb 10th 2010, 05:55 PM
BugzLooney
First and second derivative of e^6e^x?
derivative of e^6e^x?
• Feb 10th 2010, 05:56 PM
General
Quote:

Originally Posted by BugzLooney
derivative of e^6e^x?

Is it $\displaystyle e^{6e^x}$ ?
• Feb 10th 2010, 07:14 PM
BugzLooney
Yes that is the question indeed.
• Feb 10th 2010, 07:20 PM
danielomalmsteen
The first derivate, using the chain rule

$\displaystyle f(x)=e^{6e^x}$
$\displaystyle f'(x)=e^{6e^x} \cdot (6e^x)'$
$\displaystyle f'(x)=6e^{6e^x} \cdot (e^x)'$
$\displaystyle f'(x)=6e^{6e^x+x}$
• Feb 10th 2010, 07:22 PM
Pulock2009
consider the expression equal to say t. take logs on both sides of the equation then differentiate and u should get the derivative of t ie. the derivative of your expression.
• Feb 10th 2010, 07:23 PM
Pulock2009
for the second derivative differetiate again as you already might be knowing!
• Feb 10th 2010, 07:23 PM
danielomalmsteen
The second derivate

$\displaystyle f'(x)=6e^{6e^x+x}$
$\displaystyle f''(x)=6e^{6e^x+x} \cdot (6e^x+x)'$
$\displaystyle f''(x)=6e^{6e^x+x} \cdot (6e^x+1)$
• Feb 10th 2010, 07:54 PM
General
Quote:

Originally Posted by BugzLooney
Yes that is the question indeed.

If $\displaystyle g$ is a differentiable function and $\displaystyle f(x)=e^{g(x)}$, then $\displaystyle f$ is differentiable and:
$\displaystyle f'(x)=e^{g(x)} g'(x)$.
Just apply this on your problem.