Rules of differentiation

For $f(x) = x^2$ you have already worked the limit out by hand and got $f'(x) = 2x$. Repeating that for every new function would be endless busywork. Luckily nobody has to: the limit has been worked out once and for all for whole families of functions. What remains are three short rules that let you differentiate an expression like $4x^3 - 5x^2 + 6x - 9$ in a single line.

The power rule

The most important rule covers all powers $x^n$. It says: the exponent moves to the front as a factor, and the exponent itself drops by 1. So $x^3$ becomes $3x^2$, $x^5$ becomes $5x^4$, and $x^2$ becomes $2x^1$, that is $2x$. This matches exactly the result you already know for the parabola. Even plain $x$ follows the rule, because $x = x^1$ turns into $1 \cdot x^0 = 1$: the line $y = x$ has slope 1 everywhere.

Constant factors and sums

If a number sits in front of the power, it simply stays there while you differentiate. For $f(x) = 5x^3$ you differentiate $x^3$ to $3x^2$ and keep the 5, which gives $f'(x) = 15x^2$. That is the constant factor rule, and it makes visual sense: a graph stretched five times as tall runs five times as steeply at every point. If the function is made of several terms, the sum rule helps: you differentiate each term on its own and put the results back together with plus or minus. So $x^3 + x^2$ becomes $3x^2 + 2x$.

That leaves the numbers without an $x$. The graph of $f(x) = 7$ is a horizontal line at height 7, it rises nowhere and falls nowhere. Its slope is 0 everywhere, so $f'(x) = 0$. That is why the 7 in $x^2 + 7$ disappears when you differentiate: the 7 only shifts the parabola 7 units upwards, and shifting never changes the steepness at any point. So $x^2 + 100$ also has the derivative $2x$.

A polynomial step by step

Now everything together on $f(x) = 4x^3 - 5x^2 + 6x - 9$. Term by term: $4x^3$ becomes $4 \cdot 3x^2 = 12x^2$, $-5x^2$ becomes $-5 \cdot 2x = -10x$, $6x$ becomes $6$, and the $-9$ drops out. Put back together this gives $f'(x) = 12x^2 - 10x + 6$. Notice how the degree falls by 1: a function of degree three turns into a derivative of degree two. From $f'$ you can then read off every individual slope: substitute a position and you get the slope exactly there. The other way round, $f'(x) = 0$ finds the positions where the graph runs flat.