Solving Y=\sqrt{x}(x-1) Derivitive

[cscott]

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

 

[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).

 

[cscott]

Can I get any further that here?

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

 

[Jameson]

You're making it more complicated than necessary.

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.

 

[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?

 

[cscott]

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)