There are just four simple facts which suffice to take the derivative of any polynomial, and actually of somewhat more general things.

First, there is the rule for taking the derivative of a **power function** which takes the $n$th power of its input. That is,
these functions are functions of the form $f(x)=x^n$. The formula is
$${d\over dx}x^n=n\,x^{n-1}$$
That is, the exponent comes down to become a coefficient in front of
the thing, and the exponent is decreased by $1$.

The second rule, which is really a special case of this power-function
rule, is that * derivatives of constants are zero*:
$${d\over dx}\,c=0$$
for any constant $c$.

The third thing, which reflects the innocuous role of constants in
calculus, is that for *any* functions $f$ of $x$
$$\;\;{d\over dx}\,c\cdot f=c\cdot\,{d\over dx}\,f$$
The fourth is that for *any* two functions $f,g$ of $x$, *the
derivative of the sum is the sum of the derivatives*:
$${d\over dx}(f+g)={d\over dx}f+{d\over dx}g$$

Putting these four things together, we can write general formulas like $${d\over dx}(ax^m+bx^n+cx^p)=a\cdot mx^{m-1}+b\cdot nx^{n-1}+ c\cdot px^{p-1}$$ and so on, with more summands than just the three, if so desired. And in any case here are some examples with numbers instead of letters: \begin{align*} {d\over dx}5x^3&=5\cdot 3x^{3-1}=15x^2\\ {d\over dx}(3x^7+5x^3-11)&= 3\cdot7x^6+5\cdot 3x^2-0=21x^6+15x^2\\ {d\over dx}(2-3x^2-2x^3)&=0-3\cdot 2x-2\cdot 3x^2=-6x-6x^2\\ {d\over dx}(-x^4+2x^5+ 1)&=-4x^3+2\cdot 5x^4+0=-4x^3+10x^4 \end{align*}

Even if you do catch on to this idea right away, it is wise to
practice the * technique* so that not only can you do it * in
principle*, but also * in practice*.

#### Exercises

- Find ${ d \over dx }(3x^7+5x^3-11)$
- Find ${ d \over dx }(x^2+5x^3+2 )$
- Find ${ d \over dx }(-x^4+2x^5+ 1)$
- Find ${ d \over dx }(-3x^2-x^3-11)$