Q:

I think I solved it, but can someone check my solutions please? - Instantaneous Rate of Change

It's been awhile since I've done anything with rates of change and I'm struggling with deriving a formula in terms of 'x.'
From what I can recall.. the average rate of change and instantaneous rate of change are basically the same? Instantaneous is just within a smaller interval?

EDIT:
I think I'm on the right track now. I use the difference quotient I think?
I'm working through them now, but if someone could possibly verify that I'm doing them correctly, that'd be awesome! Thanks.

For example, if I'm asked to find the instantaneous rate of change of the following functions for any value of x:
1.)  $f(x) = a(x^2) + bx + c$
Solution: $ F(x) = 2ax + b $

2.) $ g(x) = \sqrt{x} $
Solution: $ G(x) = \cfrac{1}{2\sqrt{x}}$

3.)  h(x) = 1/x
Solution: h(x) = 1

I don't even get how or where to start solving any of these to get a formula for the instantaneous rate of change for x. Would I substitute x into the formula for h?
Thanks

A:

The instantaneous rate of change formula is simply the derivative which can be obtained either by going through the difference quotient calculation or simply by following standard derivative rules. For example for the third question we have
$\lim_{h\to0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}=\lim_{h\to0}\frac{\frac{x-(x+h)}{x(x+h)}}{h}=\lim_{h\to0}\frac{-1}{x(x+h)}=\frac{-1}{x^{2}}$.

