# Need derivative help once more.

• Mar 2nd 2010, 05:43 PM
Zanderist
Need 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 PM
Algebraicgeometry421
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 PM
skeeter
Quote:

Originally Posted by Zanderist
$\displaystyle f(x)= x^\frac{6}{9}$

I need help with fractional exponents.

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

first, note that $\displaystyle \frac{6}{9} = \frac{2}{3}$

$\displaystyle f(x) = x^{\frac{2}{3}}$

$\displaystyle f'(x) = \frac{2}{3} x^{\frac{2}{3} - 1} =$ ?
• Mar 2nd 2010, 05:55 PM
Zanderist
Quote:

Originally Posted by skeeter
first, note that $\displaystyle \frac{6}{9} = \frac{2}{3}$

$\displaystyle f(x) = x^{\frac{2}{3}}$

$\displaystyle f'(x) = \frac{2}{3} x^{\frac{2}{3} - 1} =$ ?

$\displaystyle f'(x) = \frac{2}{3} x^{-\frac{1}{3}}$

I have another one.

$\displaystyle 4\sqrt{t}+\frac{7}{\sqrt{t}}$
• Mar 2nd 2010, 06:17 PM
dani
Quote:

Originally Posted by Zanderist
$\displaystyle 4\sqrt{t}+\frac{7}{\sqrt{t}}$

First rewrite the equation: $\displaystyle 4t^{\frac{1}{2}}+7t^{\frac{-1}{2}}$
Now follow your rules $\displaystyle nx^{n-1}$
• Mar 2nd 2010, 07:06 PM
Zanderist
Quote:

Originally Posted by dani
First rewrite the equation: $\displaystyle 4t^{\frac{1}{2}}+7t^{\frac{-1}{2}}$
Now follow your rules $\displaystyle nx^{n-1}$

So...this..
$\displaystyle 4/2t^{\frac{1}{2}-1}+7/2t^{\frac{-1}{2}-1}$

to this...

$\displaystyle 2t^{-\frac{1}{2}}+\frac{7}{2}t^{-\frac{3}{2}}$
• Mar 2nd 2010, 07:10 PM
Zanderist
Okay let's take it up a level...

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

Quotient rule I believe.

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

It's derivative is...

$\displaystyle f '(x)= e^x \log e$
• Mar 2nd 2010, 07:55 PM
Zanderist
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 AM
skeeter
start a new problem with a new thread, otherwise this one can get very confusing.

thank you.