# 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

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"The instantaneous rate of change" What do you mean by instantaneous, can you express this in terms of a limit? Also what is changing? And with respect to what? –  Ethan May 1 '13 at 5:04
I didn't create a "new" account. That was my old one. And it's no longer allowed to ask questions on Stackoverflow because I asked a question that got downrated extremely fast by a lot of people. It's dumb but instead of using multiple accounts, I use this now. Sorry I realized it after I asked the question. –  Cozen May 1 '13 at 5:06
@Ethan that's exactly my question. I have no Idea. That is quite literally what my review question is asking me. –  Cozen May 1 '13 at 5:07
I think you should review the definition of the derivative. –  Jeremy May 1 '13 at 5:10
Look up differential calculus on Wikipedia or in a high school textbook. The right answer for 1 is $f'(x)=2ax+b$ –  Dale M May 1 '13 at 5:46
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$\lim_{h\to0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}=\lim_{h\to0}\frac{\frac{x-(x+h)}{x(x+h)}}{h}=\lim_{h\to0}\frac{-1}{x(x+h)}=\frac{-1}{x^{2}}$.
You need a common denominator to add fractions so for the above case we have $\frac{1}{x+h}-\frac{1}{x}=\frac{x}{x(x+h)}-\frac{x+h}{x(x+h)}=\frac{x-(x+h)}{x(‌​x+h)}$ –  user71352 May 1 '13 at 7:36
Note that the numerator in my previous comment is $x-(x+h)=-h$. This will cancel with h in the denominator. –  user71352 May 1 '13 at 7:37