## Content

### Some derivatives

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

 $$f(x)$$ $$f'(x)$$ constant $$c$$ 0 $$ax+b$$ $$a$$ $$x^2$$ $$2x$$ $$x^3$$ $$3x^2$$

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.

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