Examples of Derivatives

Example 1.

$y=f(x)=x.$

$$\frac{f(x+h)-f(x)}{h}=\frac{(x+h)-x}{h}=\frac{h}{h}=1.$$

As $h⟶0$, this is unaffected, hence is regarded as approaching 1 (since it is near 1 when $h$ is near zero). Thus we have

y' = f'(x) = 1\quad \text{ for all } x.

Example 2.

$y=f(x)=c$,
i.e., $f(x)$ is a constant $c$ for all $x$. (If you prefer to see an "$x$" in a function of $x$, you may write $f(x)=(x+c)-x.)$

$$\frac{f(x+h)-f(x)}{h}=\frac{c-c}{h}=0,$$

which is unchanged as $h$ approaches 0. Thus f'(x) = 0 in this case for all $x.$

Example 3.

$y=f(x)=x^n,$ $n$ a positive integer.

\begin{aligned} \frac{f(x+h)- f(x)}{h}& = \frac{(x+h)^n - x^n}{h}\\ &= \frac{x^n + nx^{n-1}h + \frac{n(n-1)}{2} x^{n-2} h^{2} + \dots + nxh^{n-1} + h^n - x^n}{h}\\ &= nx^{n-1} + \frac{n(n-1)}{2} x^{n-2} h + \dots + nxh^{n-2} + h^{n-1} . \end{aligned}

As $h\to 0$, all terms but the first go to zero, and their sum likewise goes to zero. (This seems quite obvious, but we shall eventually prove it.) Thus f'(x) = nx^{n-1}.

Example 4.

$y=f(x)=x^{p/q}$ (this is only defined for $x\ge 0$.)

$$\frac{f(x+h)-f(x)}{h}=\frac{(x+h{)}^{p/q}-x^{p/q}}{h}.$$

Let $u=(x+h{)}^{1/q},v=x^{1/q}$. Then as $h\to 0,u\to v$. Also $x+h=u^q,x=v^q$, and $h=u^q-v^q$. The difference quotient becomes

$$\frac{u^p-v^p}{u^q-v^q}=\frac{(u-v)(u^{p-1}+u^{p-2}v+\cdots +u{v}^{p-2}+v^{p-1})}{(u-v)(u^{q-1}+u^{q-2}v+\cdots +u{v}^{q-2}+v^{q-1})}$$$$=\frac{u^{p-1}+u^{p-2}v+\cdots +u{v}^{p-2}+v^{p-1}}{u^{q-1}+u^{q-2}v+\cdots +u{v}^{q-2}+v^{q-1}}.$$

The numerator consists of $p$ terms, each approaching $v^{p-1}$, and the denominator consists of $q$ terms, each approaching $v^{q-1}$. The quotient then approaches

$$\frac{p{v}^{p-1}}{q{v}^{q-1}}=\frac{p}{q}{v}^{p-q}=\frac{p}{q}(x^{1/q}{)}^{p-q}=\frac{p}{q}{x}^{(p-q)/q}=\frac{p}{q}{x}^{\frac{p}{q}-1}.$$

Here we have assumed $x\ne 0$, and that the limit of a quotient is the quotient of the limits if the denominator does not approach zero. Thus

f'(x) = \frac{p}{q} x^{\frac{p}{q}-1} ,

and we have proved that if $f(x)=x^a$, $a$ a positive rational number, i.e., a quotient of positive integers, then

f'(x)=a x^{a-1}.