The derivative of a power of $x$ is very simple. This rule is used so often that it even has its own nickname: Newton's Hammer.

Power Law (Newton's Hammer)


If $n$ is a constant, then $ \displaystyle{ \frac{d\, x^n}{dx} = (x^n)' = n x^{n-1}.} $

Of course, for this rule to make sense the function $f(x)=x^n$ has to make sense! If $n$ is a whole number, there is no problem. If $n$ is negative, then we need $x \ne 0$. If $n$ is an irrational number, or a fraction with an even denominator, then we need $x \ge 0$. Expressions like $0^{-1}$ or $(-2)^{1/2}$ aren't defined. The following video, and the text below it, shows why Newton's Hammer works when $n$ is a whole number. Later on we'll see why it works when $n$ is negative (via the quotient rule), when $n$ is a fraction (via the chain rule) and when $n$ is irrational (via exponentials and logs).

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Before we talk about general values of $n$, let's look at the derivative of $x^3$. This is \begin{eqnarray*} (x^3)' & = & \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h} \cr & = & \lim_{h \to 0} \frac{(x^3 + 3x^2h+3xh^2+h^3)-x^3}{h} \cr & = & \lim_{h \to 0} \frac{3x^2 h + 3xh^2 + h^3}{h} \cr & = & \lim_{h \to 0} 3x^2 + 3xh + h^2 \cr & = & 3x^2. \end{eqnarray*} The details of the expansion of $(x+h)^3$ aren't important. What's important is that $(x+h)^3 = x^3 + 3x^2 h + O(h^2)$, where $O(h^2)$ means "terms with 2 or more powers of $h$''. We then have \begin{eqnarray*} (x^3)' & = & \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h} \cr & = & \lim_{h \to 0} \frac{3x^2 h + O(h^2)}{h} \cr & = & \lim_{h \to 0} 3x^2 + O(h^1) = 3x^2. \end{eqnarray*} We also have $(x+h)^4 = x^4 + 4x^3h + O(h^2)$, as you can see by multiplying $(x+h)^3 = x^3 + 3x^2 h + O(h^2)$ by $x+h$, and $(x+h)^5 = x^5 + 5x^4h + O(h^2)$, as you can see by multiplying $(x+h)^4 = x^4 + 4x^3 h + O(h^2)$ by $x+h$, and so on. In general, $$ (x+h)^n = x^n + nx^{n-1} h + O(h^2), $$ so \begin{eqnarray*} (x^n)' & = & \lim_{h \to 0} \frac{(x+h)^n - x^n}{h} \cr & = & \lim_{h \to 0} \frac{nx^{n-1} h + O(h^2)}{h} \cr & = & \lim_{h \to 0} nx^{n-1} + O(h^1) = nx^{n-1}. \end{eqnarray*} There is an important special case of Newton's Hammer. When $n=0$, $x^n=x^0=1$. By definition, constants don't change, so the derivative of $x^0$ is 0. This agrees with Newton's Hammer, since $0 x^{-1}=0$.