Sums, Products, Reciprocals

Let $f(x)$ and $g(x)$ be functions having derivatives f'(x) and g'(x). Let us look for the derivative of the function $f(x)+g(x)$. The appropriate difference quotient is

\begin{aligned} &\frac{(f(x+h) + g(x+h)) - (f(x) + g(x))}{h}\\ &\qquad = \frac{f(x+h) - f(x)}{h} + \frac{g(x+h) - g(x)}{h} . \end{aligned}

As $h\to 0$, the first term goes to f'(x), the second to g'(x), and the sum to f'(x) + g'(x). This gives the rule

\bbox[5px,border:1px solid black;background-color:#f2f2f2]{(f(x) + g(x))' = f'(x) + g'(x).}\tag{*}

Now let $c$ be a number, and form the function $cf(x)$. Its difference quotient is

$$\frac{cf(x+h)-cf(x)}{h}=c\cdot \frac{f(x+h)-f(x)}{h}.$$

As $h\to 0$, this approaches cf'(x), giving the rule

\bbox[5px,border:1px solid black;background-color:#f2f2f2]{(cf(x))' = cf'(x)}.\tag{**}

On the basis of the rules (*) and (**) and the examples (b) and (c), we are ready to calculate the derivative of any polynomial, i.e., any function of the form $a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0$, where $a_0,a_1,\dots ,a_n$ are numbers and $n$ is a positive integer. For by b), the derivative of $a_0$ is 0, and by c) and (**), that of $a_i{x}^i$ is $i{a}_i{x}^{i-1}$ for $1\le i\le n$. Thus we can find the derivative of each term in the sum. From (*) we see that the derivative of a sum of two terms is the sum of their derivatives. It follows for three terms, since

\begin{aligned} \left(f(x) + g(x) + h(x)\right)' &= \left(\left(f(x) + g(x)\right) + h(x)\right)'\\ &= (f(x) + g(x))' + h'(x) \\ &= (f'(x) + g'(x)) + h'(x)\\ &= f'(x) + g'(x) + h'(x) . \end{aligned}

Note that we have simply used (*) twice. A repetition shows that the rule holds for four terms, and so on for any desired number. This is an occasion for the use of the principle of induction, about which we shall say more in the next section. We have therefore shown that if

$$f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_o,$$

then

f'(x) = na_n x^{n-1} + (n-1)a_{n-1} x^{n-2} + \dots + 2 a_2 x + a_1 .

Thus we can differentiate any polynomial.

Now consider the function $f(x)g(x)$. Its difference quotient is

$$\frac{f(x+h)g(x+h)-f(x)g(x)}{h}$$

and there is no obvious direct way of attacking this. We must therefore resort to somewhat indirect means, introducing a trick which is often useful. If the difference quotient were

$$\frac{f(x+h)g(x+h)-f(x+h)g(x)}{h}=f(x+h)\cdot \frac{g(x+h)-g(x)}{h},$$

this would at least contain an expression whose limit we recognize. But if we had used this in place of the original difference quotient, we should have had to correct it by adding their difference, or

\begin{array}{ccccc} \dfrac{f(x+h)g(x+h)-f(x)g(x)}{h} & = & f(x+h) & \dfrac{g(x+h)-g(x)}{h} & + & \dfrac{f(x+h)-f(x)}{h} & g(x) \\ & & \downarrow & \downarrow & & \downarrow & \downarrow \\ & & f(x) & g'(x) & & f'(x) & g(x) \end{array}

But as $h\to 0$, the quantities involved approach the expressions indicated by arrows below them. The only questionable matter here may be $f(x+h)\to f(x)$; this expresses a property called continuity which is possessed by all functions which have derivatives, and by many others as well. This will be discussed more fully in the next paragraph. Thus the whole expression approaches f(x)g'(x) + f'(x)g(x), and we have the rule

\bbox[5px,border:1px solid black;background-color:#f2f2f2]{(f(x)g(x))' = f(x)g'(x) + g(x)f'(x).}\tag{***}

A function $f(x)$ is called continuous at $x_0$ if whenever $x$ is near $x_0$, then $f(x)$ is near $f(x_0)$, or in other words, if

$$\underset{h\to 0}{lim}f(x_0+h)=f(x_0).$$

Now suppose f'(x_0) exists. Then

\frac{f(x_0+h) - f(x_0)}{h} - f'(x_0) = \epsilon(h) ,

where $ϵ(h)\to 0$ as $h\to 0$, and

\begin{array}{ccccccc} f(x_0 + h) - f(x_0) & = & h & f'(x_0) & + & h & \epsilon(h). \\ & & \downarrow & \downarrow & & \downarrow & \downarrow \\ & & 0 & f'(x_0) & & 0 & 0 \end{array}

As $h\to 0$, the arrows indicate the limits on the right-hand side. Thus

$$\underset{h\to 0}{lim}(f(x_0+h)-f(x_0))=0,$$

or

$$\underset{h\to 0}{lim}f(x_0+h)=f(x_0).$$

Therefore:

Next consider the function $\frac{1}{f(x)}$, where $f(x)$ is never zero. The difference quotient is

\begin{array}{l} \dfrac{\dfrac{1}{f(x+h)} - \dfrac{1}{f(x)}}{h} = \dfrac{\dfrac{f(x)-f(x+h)}{f(x+h)\cdot f(x)}}{h} \\[9pt] = -\dfrac{f(x+h)-f(x)}{h} \cdot \dfrac{1}{f(x+h)} \cdot \dfrac{1}{f(x)} \\ \phantom{{}={}-{}} \phantom{xxxx}\downarrow \phantom{xxxxxxxxxx} \downarrow \phantom{xxxxx} \downarrow \\ \phantom{{}={}-{}} \phantom{xxx}f'(x) \phantom{xxxxxxxx} \dfrac{1}{f(x)} \phantom{xxx} \dfrac{1}{f(x)} \end{array}

As $h\to 0$, this approaches

- \frac{f'(x)}{\left(f(x)\right)^2} ,

in the manner indicated by the arrows. We have therefore derived

\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\left(\frac{1}{f(x)}\right)' = - \frac{f'(x)}{\left(f(x)\right)^2}.}\tag{****}

It follows that

\begin{aligned} \left(\frac{f(x)}{g(x)}\right)' &= \left( f(x) \cdot \frac{1}{g(x)} \right)' \\[6pt] &= f(x) \cdot \left( - \frac{g'(x)}{(g(x))^2} \right) + f'(x) \cdot \frac{1}{g(x)} , \end{aligned}

or

\left( \frac{f(x)}{g(x)} \right)' = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2} .