$\displaystyle f(x)= x^\frac{6}{9}$

I need help with fractional exponents.

is it still $\displaystyle nx^n-1$

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- Mar 2nd 2010, 05:43 PMZanderistNeed derivative help once more.
$\displaystyle f(x)= x^\frac{6}{9}$

I need help with fractional exponents.

is it still $\displaystyle nx^n-1$ - Mar 2nd 2010, 05:48 PMAlgebraicgeometry421
Yes , if y=x^n then y' = n x^(n-1).

So, in your problem , y'=(6/9)(x^((6/9)-1))= (2/3)(x^(-1/3))=2/(3x^(1/3)). - Mar 2nd 2010, 05:48 PMskeeter
- Mar 2nd 2010, 05:55 PMZanderist
- Mar 2nd 2010, 06:17 PMdani
- Mar 2nd 2010, 07:06 PMZanderist
- Mar 2nd 2010, 07:10 PMZanderist
Okay let's take it up a level...

$\displaystyle \frac{e^x}{(2-1x)}$

Quotient rule I believe.

I just want to make sure about this

$\displaystyle f(x)=e^x$

It's derivative is...

$\displaystyle f '(x)= e^x \log e $ - Mar 2nd 2010, 07:55 PMZanderist
I have a problem I just can't stand.

http://www.mathhelpforum.com/math-he...1&d=1267592067

I take it to find B you need the second der. for A you just leave it as is.

I just don't know what is the proper answer. - Mar 3rd 2010, 04:24 AMskeeter
start a new problem with a new thread, otherwise this one can get very confusing.

thank you.