One differentiation question and one integration question.

Jul 2009
13
0
Texas, USA
help with problem

\(\displaystyle y = {e^x}{ln{x}} \)

Find

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

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

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Dec 2007
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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.
 
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Jul 2009
13
0
Texas, USA
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?
 

pickslides

MHF Helper
Sep 2008
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Melbourne
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 \(\displaystyle \int\frac {1-\ln{x}}{x\ln{x}} ~dx = \int\frac{1}{x}\left(\frac {1-\ln{x}}{\ln{x}}\right) ~dx =\int \frac{1-u}{u}~du\)
 
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Jul 2009
13
0
Texas, USA
yea ive got that far.. but just stuck and dont know where togo after that. :/
 

skeeter

MHF Helper
Jun 2008
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North Texas
\(\displaystyle \frac{1-u}{u} = \frac{1}{u} - 1\)
 
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