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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?

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You will find this helpful. – Michael Albanese Aug 24 '14 at 19:55
@michael Ah ha, see what you can find when you know where to look, thank you now I should be able to write up my questions a bit clearer – Paul Aug 24 '14 at 19:59
For the math formatting, replace the "\sqrt(x-3)":$$\sqrt(x-3)$$with "\sqrt{x-3}":$$\sqrt{x-3}$$ – Akiva Weinberger Aug 24 '14 at 19:59

3 Answers 3

up vote 2 down vote accepted

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.

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$$\frac{dy}{dx}=\frac{d}{dx} \left ( \frac{3x}{(x-3)^{\frac{1}{2}}} \right)=\frac{(3x)'(x-3)^{\frac{1}{2}}-3x \cdot ((x-3)^{\frac{1}{2}})'}{x-3}=\frac{3(x-3)^{\frac{1}{2}}-\frac{3}{2}x(x-3)^{\frac{1}{2}-1}}{x-3}=\frac{3(x-3)^{\frac{1}{2}}-\frac{3}{2}x(x-3)^{-\frac{1}{2}}}{x-3}=\frac{2(x-3)^{\frac{1}{2}} \cdot (3(x-3)^{\frac{1}{2}}-\frac{3}{2}x(x-3)^{-\frac{1}{2}}) }{2(x-3)^{\frac{1}{2}} \cdot (x-3) }=\frac{6(x-3)-3x}{2(x-3)^{\frac{3}{2}}}= \\ =\frac{3x-18}{2(x-3)^{\frac{3}{2}}}=\frac{3(x-6)}{2(x-3)^{\frac{3}{2}}}$$

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Maybe the answers are the same.

How about you multiply the fraction terms by $(x-3)^{1/2}$

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