<s>[INST] 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?
 [/INST] $$\frac{dy}{dx}=\frac{d}{dx} \left ( \frac{3x}{(x-3)^{\frac{1}{2}}} \right)=\frac{(3x)'(x-3)^{\frac{1}{2}}-3x \cdot ((x-3)^{\frac{1}{2}})'}{x-3}=\frac{3(x-3)^{\frac{1}{2}}-\frac{3}{2}x(x-3)^{\frac{1}{2}-1}}{x-3}=\frac{3(x-3)^{\frac{1}{2}}-\frac{3}{2}x(x-3)^{-\frac{1}{2}}}{x-3}=\frac{2(x-3)^{\frac{1}{2}} \cdot (3(x-3)^{\frac{1}{2}}-\frac{3}{2}x(x-3)^{-\frac{1}{2}}) }{2(x-3)^{\frac{1}{2}} \cdot (x-3) }=\frac{6(x-3)-3x}{2(x-3)^{\frac{3}{2}}}= \\ =\frac{3x-18}{2(x-3)^{\frac{3}{2}}}=\frac{3(x-6)}{2(x-3)^{\frac{3}{2}}}$$
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