# One differentiation question and one integration question.

• Jun 20th 2010, 11:49 PM
Beachball
One differentiation question and one integration question.
help with problem

$\displaystyle y = {e^x}{ln{x}}$

Find

$\displaystyle a) \frac {dy}{dx}$

$\displaystyle b) \int_{e}^{e^2} \frac {1-ln{x}}{xln{x}} dx$
• Jun 20th 2010, 11:57 PM
mr fantastic
Quote:

Originally Posted by Beachball
help with problem

$\displaystyle y = {e^x}{ln{x}}$

Find

$\displaystyle a) \frac {dy}{dx}$

$\displaystyle b) \int_{e}^{e^2} \frac {1-ln{x}}{xln{x}} dx$

a) Use the product rule.

b) Substitute $\displaystyle u = \ln x$.

If you need more help, please show all your work and say where you're stuck.
• Jul 21st 2010, 05:41 PM
Beachball
im still having trouble with it.. ive tried the substitution but i couldnt quite get it.. would someone be able to post the step by step result?
• Jul 21st 2010, 06:00 PM
pickslides
Here's a kick off for you

$\displaystyle \int_{e}^{e^2} \frac {1-\ln{x}}{x\ln{x}} ~dx$

$\displaystyle u = \ln{x} \implies \frac{du}{dx}= \frac{1}{x}$

Now $\displaystyle \int\frac {1-\ln{x}}{x\ln{x}} ~dx = \int\frac{1}{x}\left(\frac {1-\ln{x}}{\ln{x}}\right) ~dx =\int \frac{1-u}{u}~du$
• Jul 21st 2010, 06:02 PM
Beachball
yea ive got that far.. but just stuck and dont know where togo after that. :/
• Jul 21st 2010, 06:12 PM
skeeter
$\displaystyle \frac{1-u}{u} = \frac{1}{u} - 1$
• Jul 21st 2010, 06:20 PM
Beachball
thankyou all so much