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 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.

 

[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)