# Homework Help: Defferentiate with respect to x

1. May 23, 2010

### templarbaza

2/x^(3) + 5x^(2/3)

ok now i understand that this is what you do at the start,

2x^(-3) + 5x^(2/3)
Then Dy/Dx

-6/x^(4) + (2/3)5x^(-1/3)

After that i don't know what do do, the first part -6/x^(4) is fine whilst (2/3)5x^(-1/3) is meant to cancel down to

10/3*root(X) but i got lost around here.

2. May 23, 2010

### HallsofIvy

What exactly is your difficulty? I assume you know that (2/3)5= 10/3 so surely that is not the problem. Also you know that $x^{-1/3}= 1/x^{1/3}$ because you used that property to write $2/x^3$ as $2x^{-3}$.

The only thing left is that $x^{1/3}= \sqrt[3]{x}$.

$$(2/3)5 x^{-1/3}= \frac{10}{3\sqrt[3]{x}}$$

The "root x" is a cube root, not a square root.

3. May 23, 2010

### templarbaza

HallsofIvy Thanks for the help i understand it a lot better now ^_^

Last edited: May 23, 2010
4. May 24, 2010

### Redbelly98

Staff Emeritus
Moderator's note: thread moved from Differential Equations. In the future please post homework questions, or any textbook-style questions that are for independent study, in one of the Homework & Coursework Questions forums.