Need derivative help once more.

• Mar 2nd 2010, 06:43 PM
Zanderist
Need derivative help once more.
$f(x)= x^\frac{6}{9}$

I need help with fractional exponents.

is it still $nx^n-1$
• Mar 2nd 2010, 06:48 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, 06:48 PM
skeeter
Quote:

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

I need help with fractional exponents.

is it still $nx^n-1$

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

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

$f'(x) = \frac{2}{3} x^{\frac{2}{3} - 1} =$ ?
• Mar 2nd 2010, 06:55 PM
Zanderist
Quote:

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

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

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

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

I have another one.

$4\sqrt{t}+\frac{7}{\sqrt{t}}$
• Mar 2nd 2010, 07:17 PM
dani
Quote:

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

First rewrite the equation: $4t^{\frac{1}{2}}+7t^{\frac{-1}{2}}$
Now follow your rules $nx^{n-1}$
• Mar 2nd 2010, 08:06 PM
Zanderist
Quote:

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

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

to this...

$2t^{-\frac{1}{2}}+\frac{7}{2}t^{-\frac{3}{2}}$
• Mar 2nd 2010, 08:10 PM
Zanderist
Okay let's take it up a level...

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

Quotient rule I believe.

$f(x)=e^x$

It's derivative is...

$f '(x)= e^x \log e$
• Mar 2nd 2010, 08:55 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, 05:24 AM
skeeter
start a new problem with a new thread, otherwise this one can get very confusing.

thank you.