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Given that $\frac{d}{dx}(\frac{1}{x-3})=-\frac{1}{(x-3)^2}$, calculate the derivative of $\frac{x}{x-3}$

It looks like i need the quotient rule (which I have not learned), but since it gave the extra information there must be a quicker way of doing it.

I tried doing:$\frac{x-1}{x-3}+\frac{1}{(x-3)}$ and finding the derivative of each but it does not work. So can someone please help to calculate the derivative of $\frac{x}{x-3}$ with the given information and without the quotient rule? Thanks.

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Hint:

$$\frac{x}{x-3}=\frac{(x-3)+3}{x-3}$$

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  • $\begingroup$ Hey thanks for the hint really helped can you please check if i am allow to do this?$\frac{d}{dx}(1)+3\frac{d}{dx}(\frac{1}{x-3})$ then $0+3\frac{-1}{(x-3)^2}$ and finally $\frac{-3}{(x-3)^2}$ $\endgroup$
    – CountDOOKU
    Jul 23 '19 at 11:46
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    $\begingroup$ @FredWeasley Looks good to me! $\endgroup$
    – 5xum
    Jul 23 '19 at 12:01
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Use the product rule. $\frac d {dx} (x(\frac 1 {x-3}))=(x)\frac d {dx} (\frac 1 {x-3}) +\frac 1 {x-3}=\frac {-3} {(x-3)^{2}}$.

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