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.
Best Answer
Hint:
$$\frac{x}{x-3}=\frac{(x-3)+3}{x-3}$$