# Basic Derivative

1. Feb 22, 2006

### cscott

I need to find dy/dx of $$\frac{x^2 - 2x}{\sqrt{x}}$$

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

Does this look right so far?

Last edited: Feb 22, 2006
2. Feb 22, 2006

### happyg1

that's what I got

3. Feb 22, 2006

### assyrian_77

It is right, but can be simplified quite alot.

4. Feb 22, 2006

### cscott

Yeah, I was having trouble getting the simplified answer in my book so I wanted to check whether I was on the right track.

I can't get past here:

$$\frac{2x\sqrt{x} - 2\sqrt{x} - x^2 - 2x}{2x\sqrt{x}}$$

5. Feb 22, 2006

### assyrian_77

I could have made a mistake (very likely ) but i got

$$\frac{\frac{3}{2}x-1}{\sqrt{x}}$$

6. Feb 22, 2006

### cscott

Book says: $$\frac{3x - 2}{2\sqrt{x}}$$

7. Feb 22, 2006

### cscott

Is my algebra just really bad or are those different? :p

8. Feb 22, 2006

### assyrian_77

It's the same as mine.

9. Feb 22, 2006

### assyrian_77

This doesn't look right at all. What did you do to get here?

10. Feb 22, 2006

### cscott

I multiplied out the left two terms in the numerator, and stuck the term with the negative exponant in the denominator. I assume now from what you said above that I need to multiply the other terms by 2sqrt(x) if I want to do that. correct?

11. Feb 22, 2006

### cscott

woo, I got it nevermind! Thanks for the help.

12. Feb 22, 2006

### assyrian_77

You can't just take the term with the negative exponent and move it to the denominator.

$$\frac{A-\frac{B}{C}}{D}\neq\frac{A-B}{CD}$$

You have to make a common denominator for the numerator (if that made sense), i.e.

$$\frac{A-\frac{B}{C}}{D}=\frac{AC-B}{CD}$$

13. Feb 22, 2006

### assyrian_77

Ok, good thing.

14. Feb 22, 2006

### cscott

Yeah, I saw that in posts 5-6. That was my mistake. Thanks again.