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I'm having a lot of trouble understanding the core concept of finding $f\,'(a)$ or any number in place of $a$. I do not understand what finding $f\,'$ means to begin with, or how to solve them. If somebody could give any information i would be appreciative.

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Find the derivative of the function in question and evaluate it at $a$. –  Clayton Feb 14 '13 at 4:42
    
Is the problem finding the f' using the definition of the derivative –  i.a.m Feb 14 '13 at 4:57
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If you know as little as you claim to, what you need to do is sit down with an calculus textbook and start working through it from the beginning. You can ask here about trouble you have understanding specific parts of what the textbook says, but explaining all of calculus from nothing in a Math.SE answer is a bit of a tall order. –  Henning Makholm Feb 14 '13 at 5:13
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3 Answers

It's just the derivative. f'(a) is the derivative of f(x) at x=a. So for example if f(x)=$x^2$ then f'(x)=$2x$ and f'(a)=$2a$

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$f'(x)$ is the derivative of $f$ at $x$.

Other similar meanings are

rate of change

slope

grade (of an incline)

pitch, etc.

You know that slope of a line is the ratio of rise over run.

Well the graph of a function may not be a straight line but we can still find something that plays the same role as slope of a line.

Here is how, at a point with $x=a$ on the graph draw its tangent line. The slope of this line is the derivative of $f$ at $a$.

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The easiest way to find the derivative using the definition is by using this limit$$f'(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$$, For example if $f(x)=x^2-1$ find the derivative at x=1, using the definition we have $$f'(1)=\lim_{x\rightarrow 1}\frac{f(x)-f(1)}{x-1}=\lim_{x\rightarrow 1}\frac{x^2-1-0}{x-1}$$$$=\lim_{x\rightarrow 1}\frac{(x-1)(x+1)}{x-1}$$$$=\lim_{x\rightarrow 1}x+1=2.$$This is the hardest part. After you learn more about derivatives you will learn some rules which are relatively easier to use than the definition, and things should get easier from there.

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