# How to do this derivative question without using the quotient rule?

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.

$$\frac{x}{x-3}=\frac{(x-3)+3}{x-3}$$
• 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 $\frac{-3}{(x-3)^2}$ Jul 23 '19 at 11:46
Use the product rule. $$\frac d {dx} (x(\frac 1 {x-3}))=(x)\frac d {dx} (\frac 1 {x-3}) +\frac 1 {x-3}=\frac {-3} {(x-3)^{2}}$$.