Find simplified formula for the difference quotient (f(x+h)-f(x))/h

You are given the two problem
1. f(x)=1/x
2. f(x)= 2^(x)

I know that you are supposed to replace f(x) with whatever they say it is equal to, in out case. 1/x and 2^x.

For #1, would the equation become ((1/x+h) -1/x)/ h
which can then become (1/x +1/h -1/x)/h.
after cancelling out the 1/x's you are left with (1/h)/h
h/h=1
1/1= 1 Is the answer 1?
I know this is easy, but for some reason I feel like I'm doing it completely wrong.
And I don't really know where to begin with question2, but this is what I have so far.
(2^(x+h) -2^x)/h Could I simplify this any further?

Please show exactly how you solve, I'm getting so frustrated!!

Your problem with the first question is when you split up 1/(x + h) to 1/x + 1/h; you cannot do this. For example, you know that 1/2 = 1/(1 + 1), but you know that this does not equal to 1/1 + 1/1 = 2.

So, for the first one you have:
[f(x + h) - f(x)]/h = [1/(x + h) - 1/x]/h.

To clear the fractions in the numerator, multiply the numerator and denominator of the main fraction by x(x + h), the LCD of the fractions in the numerator, to get:
[1/(x + h) - 1/x]/h = [x - (x + h)]/[xh(x + h)]
= -h/[xh(x + h)]
= -1/[x(x + h)], by canceling h.

The second question cannot be simplified that much. The most you can do is factor out 2^x to get:
[2^(x + h) - 2^x]/h = 2^x(2^h - 1)/h,

but that's about it.

I hope this helps!