How do I find the derivative of: y = 2/(2-x)

I know that I will have to take the limit of "(f(x+h) - f(x))/h" as h approaches 0, but the algebra is really confusing me. If anyone could help, that would be much appreciated. Thanks in advance!

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- Jun 17th 2013, 09:39 AMmisiaizeskaFinding derivative by first principles?
How do I find the derivative of: y = 2/(2-x)

I know that I will have to take the limit of "(f(x+h) - f(x))/h" as h approaches 0, but the algebra is really confusing me. If anyone could help, that would be much appreciated. Thanks in advance! - Jun 17th 2013, 09:46 AMHallsofIvyRe: Finding derivative by first principles?
You know that $\displaystyle f(x)= \frac{2}{2- x}$ so it should be easy to see that $\displaystyle f(x+h)= \frac{2}{2- (x+h)}= \frac{2}{2- x- h}$.

Now $\displaystyle f(x+h)- f(x)= \frac{2}{2- x- h}- \frac{2}{2- x}$. You need to get "common denominators" by multiplying both numerator and denominator of the first fraction by 2- x and numerator and denominator of the second fraction by 2- x- h:

$\displaystyle \frac{2(2- x)}{(2-x-h)(2- x)}- \frac{2(2- x- h)}{(2- x-h)(2- x)}= \frac{(4- 2x)- (4- 2x- 2h)}{(2- x- h)(2- x)}$

Can you finish that?