Basic Derivative: dy/dx of \frac{x^2-2x}{\sqrt{x}}

[cscott]

I need to find dy/dx of \frac{x^2 - 2x}{\sqrt{x}}

\frac{(\sqrt{x})(2x - 2) - (x^2 - 2x)(1/2x^{-1/2})}{x}Does this look right so far?

 

[happyg1]

that's what I got

 

[assyrian_77]

It is right, but can be simplified quite alot.

 

[cscott]

Yeah, I was having trouble getting the simplified answer in my book so I wanted to check whether I was on the right track.

I can't get past here:

\frac{2x\sqrt{x} - 2\sqrt{x} - x^2 - 2x}{2x\sqrt{x}}

 

[assyrian_77]

I could have made a mistake (very likely ) but i got

\frac{\frac{3}{2}x-1}{\sqrt{x}}

 

[cscott]

Book says: \frac{3x - 2}{2\sqrt{x}}

 

[cscott]

Is my algebra just really bad or are those different? :p

 

[assyrian_77]

It's the same as mine.

 

[assyrian_77]

I can't get past here:

\frac{2x\sqrt{x} - 2\sqrt{x} - x^2 - 2x}{2x\sqrt{x}}

This doesn't look right at all. What did you do to get here?

 

[cscott]

I multiplied out the left two terms in the numerator, and stuck the term with the negative exponant in the denominator. I assume now from what you said above that I need to multiply the other terms by 2sqrt(x) if I want to do that. correct?

 

[cscott]

woo, I got it nevermind! Thanks for the help.

 

[assyrian_77]

You can't just take the term with the negative exponent and move it to the denominator.

\frac{A-\frac{B}{C}}{D}\neq\frac{A-B}{CD}

You have to make a common denominator for the numerator (if that made sense), i.e.

\frac{A-\frac{B}{C}}{D}=\frac{AC-B}{CD}

 

[assyrian_77]

woo, I got it nevermind! Thanks for the help.

Ok, good thing.

 

[cscott]

Yeah, I saw that in posts 5-6. That was my mistake. Thanks again.