Y $\displaystyle ln 5x^2$

Y' $\displaystyle 1/5x^2 . 10x = 10x/5x^2 = 10/5x = 2^-1$

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- Jul 15th 2009, 04:18 PM #1

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- Jul 15th 2009, 04:31 PM #2

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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 #3

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- Jul 15th 2009, 04:45 PM #4

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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.