# ln help

• Jul 15th 2009, 04:18 PM
crashuk
ln help
Y $\displaystyle ln 5x^2$
Y' $\displaystyle 1/5x^2 . 10x = 10x/5x^2 = 10/5x = 2^-1$
• Jul 15th 2009, 04:31 PM
pomp
Quote:

Originally Posted by crashuk
$\displaystyle ln 5x^2$
$\displaystyle 1/5x^2 . 10x = 10x/5x^2 = 10/5x = 2^-1$

Please review your post. It is almost nonsensical due to bad presentation. There is also no question.

I'm guesing what you meant to say was:

Is this correct?

$\displaystyle \frac{d}{dx} \left( \ln{\left(5x^2 \right)} \right) = 10x \times \frac{1}{5x^2} = \frac{2}{x}$?

If so, then yes it is.
• Jul 15th 2009, 04:40 PM
crashuk
yes thanks, sorry about the way i post it havent a clue to do some things on line.
• Jul 15th 2009, 04:45 PM
pomp
That's fine, just try to make sure it's at least possible to get the gist of what you're saying. Write a little, explain what it is you are trying to do, say where you are stuck and where you need help. Simply posting 2 equations means nothing.

You might find this useful:

http://www.mathhelpforum.com/math-he...-tutorial.html

It will help a lot with general mathematical formatting.