# Simple Derivitive

#### cscott

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

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#### Jameson

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

Distribute and take it away.

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

#### 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).

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#### cscott

Can I get any further that here?

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

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#### 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.

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#### cscott

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

correct?

#### Jameson

First part is incorrect, second part is correct.

#### cscott

Hmm I don't see how

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

#### 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?

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#### cscott

Jameson said:
Oh, ok. You were righting a mixed fraction. It would be best to call 1.5 $$\frac{3}{2}$$

Try not to use mixed fractions, they get too confusion. Use improper ones.
Thanks for the tip and your help (Berislav too)

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