Small part of a larger problem

[efekwulsemmay]

1. The problem statement, all variables and given/known data
I am trying to figure out of \frac{9}{x+h} can be split into some thing like
\frac{9}{x} + ?


2. Relevant equations
None


3. The attempt at a solution
I am not sure what to do. I am trying to do this as part of a larger limits problem.

[Bohrok]

I don't believe you can change one to the other. The difference between 9/(x + h) and 9/x is that the first is shifted -h units to the left of the graph of 9/x. 9/x + something would shift the graph of 9/x up something units.

[slider142]

You would need to add it to a fraction whose denominator 'a' had the property xa = x + h, or a = (x + h)/x. Unfortunately, there is no fraction that you can add that will not affect the numerator as well.

[efekwulsemmay]

Damn, ok thanks for your help, both of you.

[HallsofIvy]

What you can do is get a common denominator and subtract fractions.
\frac{9}{x+h}- \frac{9}{x}= \frac{9x}{x(x+h)}- \frac{9(x+h)}{x(x+h)}
= \frac{9x- 9(x+h)}{x(x+h)}= \frac{-9h}{x(x+h)}
and you should be able to complete the derivative.

[efekwulsemmay]

What you can do is get a common denominator and subtract fractions.
\frac{9}{x+h}- \frac{9}{x}= \frac{9x}{x(x+h)}- \frac{9(x+h)}{x(x+h)}
= \frac{9x- 9(x+h)}{x(x+h)}= \frac{-9h}{x(x+h)}
and you should be able to complete the derivative.

:surprisedAhhhhh, I see said the blind man to the deaf dog with no ears. Thank you!!!!!

Btw how did you figure out that thats what I was trying to do? That was pretty amazing. :D

[HallsofIvy]

Hey, after some time here you get used to figuring out what people are really asking!