df(x)/dx = lim(e-->0){(f(x+e) - f(x))/e}
If you think about it, the term f(x+e) - f(x) is just the change in the value of f evaluated at two nearby points and e is the interval in x over which that change occurs so the fraction {f(x+e) - f(x)}/e is the slope of a straight line defined by the values of f and the interval in x of e. The limit function draws these two points togetherso the straight line approximation now becomes exact.
Let's work one simple example of a function as a derivative using the above definition. Let f(x) = x^2 and we''l find df/dx
df/dx = lim(e-->0){(x+e)^2 -x^2)/e} = lim(x-->0){(x^2 + 2ex + e^2 -x^2)/e}
= lm(e-->0){(2ex + e^2)/e} = lim(e-->0){2x + e} now take e = 0
df/dx = 2x
You can immediately see that if f(x) = c a constant, then df/dx = 0 as it has to be since f(x) = c will not change for any value of x.
You can appy this definition to any function of f(x) that is continuous and with some care, you can also find derivatives of discontinuous functions in the intervals were thaey are continuous. Fortunately, there are rules for doing differentiation so you don't always have to go through this limit process. for instance, if f(x) = x^n where n is some number then
df/dx = n*x^(n-1)
Now your f'(1) , etc requires you to first find df/dx and then evalute the derivative for x = 1 or whatever the numberical argument is. Your example:
f(x) = 2 - 3x^2
df/dx = -6x
f'(1) = -6, f'(2) = -12 etc.