# Homework Help: Simple Derivitive

1. Jul 13, 2005

### cscott

How do I deal with the square root in $y = \sqrt{x}(x - 1)$?

2. Jul 13, 2005

### Jameson

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

Distribute and take it away.

Also, remember that $$a^x*a^y=a^{(x+y)}$$

3. Jul 13, 2005

### Berislav

You know the simple formula for deriving powers of x, right? Well, $\sqrt{x}=x^{\frac{1}{2}}$

EDIT: I was slow. Sorry, I didn't mean chain, I meant distrubution for derivation (didn't know what you call it in English).

Last edited: Jul 13, 2005
4. Jul 13, 2005

### cscott

Can I get any further that here?

$$\left(x - 1\right)\left(\frac{1}{2\sqrt{x}}\right) + \sqrt{x}$$

Last edited: Jul 13, 2005
5. Jul 13, 2005

### Jameson

You're making it more complicated than neccessary.

Distribute the $\sqrt{x}$ then take the derivative.

Ok, your way works, but I wouldn't do it that way. That's the beauty of it though, many correct ways to get the same answer.

Last edited: Jul 13, 2005
6. Jul 13, 2005

### cscott

$$1\frac{1}{2}x^{\frac{1}{2}} - \frac{1}{2}x^{-\frac{1}{2}}$$

correct?

7. Jul 13, 2005

### Jameson

First part is incorrect, second part is correct.

8. Jul 13, 2005

### cscott

Hmm I don't see how

$x^{\frac{1}{2}} \cdot x^1 = x^{1.5}$ so doesn't that become $1.5 \cdot x^{\frac{1}{2}}$?

9. Jul 13, 2005

### Jameson

Oh, ok. You were righting a mixed fraction. It would be best to write 1.5 as $\frac{3}{2}$

Try not to use mixed fractions, they get too confusing. Use improper ones.

For example: take the derivative of $$3\frac{1}{2}\frac{5}{7}x^4$$ with respect to x. Make sense?

Last edited: Jul 13, 2005
10. Jul 13, 2005

### cscott

Thanks for the tip and your help (Berislav too)