# Simple Derivitive

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

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

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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Gold Member
MHB
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?

Gold Member
MHB
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 $1.5 \cdot x^{\frac{1}{2}}$?

Gold Member
MHB
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)