## Content

### Some derivatives

So far in the examples and exercises we have found the following derivatives.

\(f(x)\) | \(f'(x)\) | |
---|---|---|

constant | \(c\) | 0 |

linear | \(ax+b\) | \(a\) |

\(x^2\) | \(2x\) | |

\(x^3\) | \(3x^2\) |

Note that the example \(f(x) = ax+b\) includes the case \(f(x) = x\), which has derivative \(f'(x) = 1\). Since we have seen that

\[ \dfrac{d}{dx} (x) = 1, \qquad \dfrac{d}{dx} (x^2) = 2x \qquad\text{and}\qquad \dfrac{d}{dx} (x^3) = 3x^2, \]it is natural to conjecture that the derivative of \(x^n\) is \(nx^{n-1}\).

We will now compute the derivative of \(f(x) = x^n\), for any positive integer \(n\). To do so, we need the binomial expansion

\[ (a+b)^n = a^n + \dbinom{n}{1} a^{n-1} b + \dbinom{n}{2} a^{n-2} b^2 + \dots + \dbinom{n}{n-2} a^2 b^{n-2} + \dbinom{n}{n-1} a b^{n-1} + b^n. \]See the module The binomial theorem for details.

We begin by using the binomial theorem to expand

\begin{align*} f(x + \Delta x) &= (x + \Delta x)^n \\ &= x^n + \dbinom{n}{1} x^{n-1} (\Delta x) + \dbinom{n}{2} x^{n-2} (\Delta x)^2 + \dots + \dbinom{n}{n-1} x (\Delta x)^{n-1} + (\Delta x)^n. \end{align*}We can then compute

\begin{align*} \dfrac{f(x+\Delta x) - f(x)}{\Delta x} &= \dfrac{(x+\Delta x)^n - x^n}{\Delta x} \\ &= \dbinom{n}{1} x^{n-1} + \dbinom{n}{2} x^{n-2} (\Delta x) + \dots + \dbinom{n}{n-1} x (\Delta x)^{n-2} + (\Delta x)^{n-1}. \end{align*}The last line has a \(\Delta x\) in every term except the first. Since \(\binom{n}{1} = n\), we have

\begin{align*} f'(x) &= \lim_{\Delta x \to 0} \dfrac{f(x + \Delta x) - f(\Delta x)}{\Delta x} \\ &= nx^{n-1}. \end{align*}The fact that the derivative of \(x^n\) is \(nx^{n-1}\) holds more generally than just when \(n\) is a positive integer. The next exercise shows it is also true when \(n=-1\). exercise 14, later in this module, shows how to find the derivative of negative integer powers of \(x\) in general.

##### Exercise 5

Prove that the derivative of \(f(x) = \dfrac{1}{x}\) is \(f'(x) = - \dfrac{1}{x^2}\). That is, prove that

\[ \dfrac{d}{dx}(x^{-1}) = -x^{-2}. \]In fact, it's also true that for any non-zero rational number (i.e., fraction) \(n\), the derivative of \(f(x) = x^n\) is \(f'(x) = nx^{n-1}\). See exercises 20 and 21 later in this module.

Even more generally, for any real number \(a\), including irrational \(a\), the derivative of

\[ f(x) = x^a \qquad\text{is}\qquad f'(x) = a x^{a-1}. \]It is not obvious how to even define what it means to raise a number to the power of an irrational number. For instance, \(2^3\) just means \(2 \times 2 \times 2\), and \(2^{\frac{7}{5}}\) just means \(\sqrt[5]{2^7}\), but what does \(2^{\sqrt{3}}\) mean? In the module Exponential and logarithmic functions, we explore these issues, show how to define \(x^a\) precisely for any real number \(a\), and show that the derivative of \(x^a\) is \(a x^{a-1}\).

In summary, the following theorem is true.

###### Theorem

For any real number \(a\), the derivative of \(f(x) = x^a\) is \(f'(x) = a x^{a-1}\), wherever \(f(x)\) is defined.