# 2 derivative problems

• November 3rd 2008, 06:19 PM
JimDavid
2 derivative problems
Quote:

Find the derivative of ln e^u/u
the answer is 0 but I don't know how to get it

lim
h-> 0 square root of (25-h) -5/h

The answer is 1/10 but I once again don't know how to do it.
• November 3rd 2008, 06:25 PM
Jameson
If it's supposed to be 0 then you must mean $y= \frac{\ln(e^u)}{u}$

$y=\frac{u\ln(e)}{u}$ by log rules

$y=\frac{u*1}{u}=1$

y'=0
• November 3rd 2008, 06:46 PM
JimDavid
Quote:

Originally Posted by Jameson
If it's supposed to be 0 then you must mean $y= \frac{\ln(e^u)}{u}$

$y=\frac{u\ln(e)}{u}$ by log rules

$y=\frac{u*1}{u}=1$

y'=0

Thank you for the help, can you help me with the second problem?
• November 3rd 2008, 06:52 PM
Jameson
If I'm reading it right it's $\lim_{h \rightarrow 0} \frac{\sqrt{25-h}-5}{h}$

Multiply the numerator and denominator by $\sqrt{25-h}+5$

$\frac{\sqrt{25-h}-5}{h} * \frac{\sqrt{25-h}+5}{\sqrt{25-h}+5}$

This comes out to be $\frac{-h}{h(\sqrt{25-h}+5)}=\frac{-1}{\sqrt{25-h}+5}$

Now try the limit again.
• November 3rd 2008, 07:00 PM
JimDavid
Quote:

Originally Posted by Jameson
If I'm reading it right it's $\lim_{h \rightarrow 0} \frac{\sqrt{25-h}-5}{h}$

Multiply the numerator and denominator by $\sqrt{25-h}+5$

$\frac{\sqrt{25-h}-5}{h} * \frac{\sqrt{25-h}+5}{\sqrt{25-h}+5}$

This comes out to be $\frac{-h}{h(\sqrt{25-h}+5)}=\frac{-1}{\sqrt{25-h}+5}$

Now try the limit again.

it came out -1/10, but the answer is 1/10

did I do something wrong there?
• November 3rd 2008, 07:03 PM
Jameson
No, I'm pretty sure the answer is negative. I just checked the limit on a calculator and it confirms it's -1/10.