# Math Help - Need derivative help once more.

1. ## Need derivative help once more.

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

I need help with fractional exponents.

is it still $nx^n-1$

2. 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)).

3. 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 $\frac{6}{9} = \frac{2}{3}$

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

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

4. 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.

$4\sqrt{t}+\frac{7}{\sqrt{t}}$

5. 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 $nx^{n-1}$

6. 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...

$2t^{-\frac{1}{2}}+\frac{7}{2}t^{-\frac{3}{2}}$

7. 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...

$f '(x)= e^x \log e$

8. I have a problem I just can't stand.

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.

9. start a new problem with a new thread, otherwise this one can get very confusing.

thank you.