Simplify the derivative of $y=\frac{3x}{\sqrt{x-3}}$ First of I'm still not totally clued up on how to format all this math properly so some of it may look a bit strange, any help in fixing it would be greatly appreciated. Anyhow onto the math...
So I have to differentiate 
$$y= \frac{3x}{\sqrt(x-3)}$$
And I used the quotient rule and now I'm at
$$\frac{3(x-3)^{1/2} - 3/2 x (x-3)^{-1/2}}{x-3}$$
The expected answer is 
$$\frac{3(x-6)}{ 2(x-3)^{3/2}}$$
So I'm slightly off, any tips?
 A: Starting with your answer and multiplying the top and bottom by $(x-3)^\frac12$ and then 2 gives:
$$
\begin{align}
\frac{3(x-3)^\frac12 - \frac32x(x-3)^{-\frac12}}{x-3} &= \frac{3(x-3) - \frac32}{(x-3)^\frac32}\\
&= \frac{6(x-3) - 3x}{2(x-3)^\frac32}\\
&= \frac{6x - 18 - 3x}{2(x-3)^\frac32}\\
&= \frac{3x - 18}{2(x-3)^\frac32}\\
&= \frac{3(x-6)}{2(x-3)^\frac32}\\
\end{align}
$$
So your answer is actually correct.
As a general rule, mathematicians don't like leaving negative powers in fractions. Multiplying top and bottom by the positive version of the power will make the negative power disappear and usually leave you with an opportunity to simplify further. The same goes for fractions-within-fractions: multiplying top and bottom of the big fraction by the bottom of the little fraction will make things much simpler.
A: $$\frac{dy}{dx}=\frac{d}{dx} \left ( \frac{3x}{(x-3)^{\frac{1}{2}}} \right)=\frac{(3x)'(x-3)^{\frac{1}{2}}-3x \cdot ((x-3)^{\frac{1}{2}})'}{x-3}=\frac{3(x-3)^{\frac{1}{2}}-\frac{3}{2}x(x-3)^{\frac{1}{2}-1}}{x-3}=\frac{3(x-3)^{\frac{1}{2}}-\frac{3}{2}x(x-3)^{-\frac{1}{2}}}{x-3}=\frac{2(x-3)^{\frac{1}{2}} \cdot (3(x-3)^{\frac{1}{2}}-\frac{3}{2}x(x-3)^{-\frac{1}{2}}) }{2(x-3)^{\frac{1}{2}} \cdot (x-3) }=\frac{6(x-3)-3x}{2(x-3)^{\frac{3}{2}}}= \\ =\frac{3x-18}{2(x-3)^{\frac{3}{2}}}=\frac{3(x-6)}{2(x-3)^{\frac{3}{2}}}$$
A: Maybe the answers are the same.
How about you multiply the fraction terms by $(x-3)^{1/2}$
