# Definition of Derivative

There is one limit that is used very frequently in the applications of calculus and different sciences. This limit has the form $\lim_{{{h}\to{0}}}\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}$ and has a special notation.

Definition. The derivative of a function $f$ is a function that is denoted by ${f{'}}{\left({x}\right)}$ and is calculated as $f{'}{\left({x}\right)}=\lim_{{{h}\to{0}}}\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}$.

The process of finding the derivative is called differentiating.

There are different notations for the derivative.

Leibniz's notation: $\frac{{{d}{y}}}{{{d}{x}}}$, $\frac{{{d}{f}}}{{{d}{x}}}$, $\frac{{d}}{{{d}{x}}}{f{{\left({x}\right)}}}$. Note that here $\frac{{{d}{y}}}{{{d}{x}}}$ is not a ratio, it is just a synonym for ${f{'}}{\left({x}\right)}.$ It is a very useful and suggestive notation, the derivative is defined as the limit of the ratio. If we want to indicate the value of the derivative in Leibniz's notation at a specific point ${a}$, we use the notation $\frac{{{d}{y}}}{{{d}{x}}}{\mid}_{{{x}={a}}}$.

Lagrange's notation: ${y}'$, ${f{'}}{\left({x}\right)}$.

Cauchy's notation: ${D}{y}$, ${D}{f{{\left({x}\right)}}}$.

In Lagrange's and Cauchy's notations, if we need to state explicitly with respect to what variable we take the derivative, we write ${y}'_{{x}}$, ${f{'}}_{{x}}$, ${D}_{{x}}{y}$, ${D}_{{x}}{f{}}$.

We will be mainly using Lagrange's notation.

Example 1. If ${f{{\left({x}\right)}}}={{x}}^{{2}}$, find ${f{'}}{\left({x}\right)}$ and ${f'{{\left({3}\right)}}}$.

We have that ${f{{\left({x}+{h}\right)}}}={{\left({x}+{h}\right)}}^{{2}}$.

Thus,

${f{'}}{\left({x}\right)}=\lim_{{{h}\to{0}}}\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}=\lim_{{{h}\to{0}}}\frac{{{{\left({x}+{h}\right)}}^{{2}}-{{\left({x}\right)}}^{{2}}}}{{{h}}}=\lim_{{{h}\to{0}}}\frac{{{{x}}^{{2}}+{2}{x}{h}+{{h}}^{{2}}-{{x}}^{{2}}}}{{h}}=$

$=\lim_{{{h}\to{0}}}\frac{{{2}{x}{h}+{{h}}^{{2}}}}{{h}}=\lim_{{{h}\to{0}}}{\left({2}{x}+{h}\right)}={2}{x}$.

So, ${f{'}}{\left({x}\right)}={2}{x}$ and ${f{'}}{\left({3}\right)}={2}\cdot{3}={6}$.

Let's work another example.

Example 2. Find ${f{'}}{\left({x}\right)}$ if ${f{{\left({x}\right)}}}=\sqrt{{{x}}}$ and state the domain of ${f{'}}$.

${f{'}}{\left({x}\right)}=\lim_{{{h}\to{0}}}\frac{{\sqrt{{{x}+{h}}}-\sqrt{{{x}}}}}{{h}}=\lim_{{{h}\to{0}}}\frac{{\sqrt{{{x}+{h}}}-\sqrt{{{x}}}}}{{h}}\frac{{\sqrt{{{x}+{h}}}+\sqrt{{{x}}}}}{{\sqrt{{{x}+{h}}}+\sqrt{{{x}}}}}=\lim_{{{h}\to{0}}}\frac{{{x}+{h}-{x}}}{{{h}{\left(\sqrt{{{x}+{h}}}+\sqrt{{{x}}}\right)}}}=$

$=\lim_{{{h}\to{0}}}\frac{{1}}{{\sqrt{{{x}+{h}}}+\sqrt{{{x}}}}}=\frac{{1}}{{\sqrt{{{x}+{0}}}+\sqrt{{{x}}}}}=\frac{{1}}{{{2}\sqrt{{{x}}}}}$

Thus, the domain of ${f{'}}$ is ${\left({0},\infty\right)}$. As can be seen, it is smaller than the domain of ${f{{\left({x}\right)}}}=\sqrt{{{x}}}$.

What if the function is given by a table of values?

Example 3. Estimate ${f{'}}{\left({1}\right)}$ and ${f{'}}{\left({1.5}\right)}$ if

 ${x}$ 1 1.5 2 ${f{{\left({x}\right)}}}$ 2 3 7

We have that ${f{'}}{\left({x}\right)}=\lim_{{{h}\to{0}}}\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}$. But if we take ${h}={0.5}$, we will obtain the following approximation: ${f{'}}{\left({x}\right)}\approx\frac{{{f{{\left({x}+{0.5}\right)}}}-{f{{\left({x}\right)}}}}}{{0.5}}={2}{\left({f{{\left({x}+{0.5}\right)}}}-{f{{\left({x}\right)}}}\right)}$.

Thus,

${f{'}}{\left({1}\right)}\approx{2}{\left({f{{\left({1}+{0.5}\right)}}}-{f{{\left({1}\right)}}}\right)}={2}{\left({f{{\left({1.5}\right)}}}-{f{{\left({1}\right)}}}\right)}={2}{\left({3}-{2}\right)}={2}$.

${f{'}}{\left({1.5}\right)}\approx{2}{\left({f{{\left({1.5}+{0.5}\right)}}}-{f{{\left({1.5}\right)}}}\right)}={2}{\left({f{{\left({2}\right)}}}-{f{{\left({1.5}\right)}}}\right)}={2}{\left({7}-{3}\right)}={8}$.

Definition. A function ${f{}}$ is differentiable at ${a}$ if ${f{'}}{\left({a}\right)}$ exists.

Definition. A function ${f{}}$ is differentiable on an open interval ${\left({a},{b}\right)}$ or ${\left({a},\infty\right)}$, or ${\left(-\infty,{a}\right)}$, or ${\left(-\infty,\infty\right)}$ if it is differentiable at every number in that interval.

Example 4. Where is ${f{{\left({x}\right)}}}={\left|{x}\right|}$ differentiable?

If ${x}>{0}$, we can say that ${\left|{x}\right|}={x}$ and we can choose an ${h}$ small enough, so that ${x}+{h}>{0}$, and, hence, ${\left|{x}+{h}\right|}={x}+{h}$.

Therefore, for ${x}>{0}$, ${f{'}}{\left({x}\right)}=\lim_{{{h}\to{0}}}\frac{{{\left|{x}+{h}\right|}-{\left|{x}\right|}}}{{h}}=\lim_{{{h}\to{0}}}\frac{{{x}+{h}-{x}}}{{h}}=\lim_{{{h}\to{0}}}{1}={1}$, and so, ${f{}}$ is differentiable for any ${x}>{0}$.

Similarly, if ${x}<{0}$, we can say that ${\left|{x}\right|}=-{x}$ and we can choose an ${h}$ small enough, so that ${x}+{h}<{0}$, and, hence, ${\left|{x}+{h}\right|}=-{\left({x}+{h}\right)}$.

Therefore, for ${x}<{0}$, ${f{'}}{\left({x}\right)}=\lim_{{{h}\to{0}}}\frac{{{\left|{x}+{h}\right|}-{\left|{x}\right|}}}{{h}}=\lim_{{{h}\to{0}}}\frac{{-{\left({x}+{h}\right)}-{\left(-{x}\right)}}}{{h}}=\lim_{{{h}\to{0}}}-{1}=-{1}$, and so, ${f{}}$ is differentiable for any ${x}<{0}$.

For ${x}={0}$, we have to investigate${f{'}}{\left({0}\right)}=\lim_{{{h}\to{0}}}\frac{{{\left|{0}+{h}\right|}-{\left|{0}\right|}}}{{h}}=\lim_{{{h}\to{0}}}\frac{{{\left|{h}\right|}}}{{h}}$.

Let's compute the one-sided limits:

$\lim_{{{h}\to{{0}}^{+}}}\frac{{\left|{h}\right|}}{{h}}=\lim_{{{h}\to{{0}}^{+}}}\frac{{h}}{{h}}=\lim_{{{h}\to{{0}}^{+}}}{1}={1}$

$\lim_{{{h}\to{{0}}^{{-}}}}\frac{{\left|{h}\right|}}{{h}}=\lim_{{{h}\to{{0}}^{{-}}}}-\frac{{h}}{{h}}=\lim_{{{h}\to{{0}}^{{-}}}}-{1}=-{1}$

Since $\lim_{{{h}\to{{0}}^{+}}}\frac{{{\left|{h}\right|}}}{{h}}\ne\lim_{{{h}\to{{0}}^{{-}}}}\frac{{{\left|{h}\right|}}}{{h}}$, we can state that $\lim_{{{h}\to{0}}}\frac{{{\left|{h}\right|}}}{{h}}$ doesn't exist and thus ${f{{\left({x}\right)}}}$ is not differentiable at ${x}={0}.$

Therefore, ${\left|{x}\right|}$ is differentiable at all values of ${x}$ except ${0}$.

Theorem. If ${f{}}$ is differentiable at ${a}$, ${f{}}$ is continuous at ${a}$. The converse is not true.

Proof. Since ${f{}}$ is differentiable at ${a}$, we can say that ${f{'}}{\left({a}\right)}=\lim_{{{h}\to{0}}}\frac{{{f{{\left({a}+{h}\right)}}}-{f{{\left({a}\right)}}}}}{{h}}$.

If we make the substitution ${x}={a}+{h}$, we have that ${h}={x}-{a}$. As ${h}\to{0}$, ${x}\to{a}$.

Therefore, ${f{'}}{\left({a}\right)}=\lim_{{{x}\to{a}}}\frac{{{f{{\left({x}\right)}}}-{f{{\left({a}\right)}}}}}{{{x}-{a}}}$.

Now, consider the identity ${f{{\left({x}\right)}}}-{f{{\left({a}\right)}}}={f{{\left({x}\right)}}}-{f{{\left({a}\right)}}}$, or, equivalently, ${f{{\left({x}\right)}}}-{f{{\left({a}\right)}}}=\frac{{{f{{\left({x}\right)}}}-{f{{\left({a}\right)}}}}}{{{x}-{a}}}{\left({x}-{a}\right)}$. Taking the limits of both sides gives $\lim_{{{x}\to{a}}}{\left({f{{\left({x}\right)}}}-{f{{\left({a}\right)}}}\right)}=\lim_{{{x}\to{a}}}{\left(\frac{{{f{{\left({x}\right)}}}-{f{{\left({a}\right)}}}}}{{{x}-{a}}}{\left({x}-{a}\right)}\right)}$.

Using the property of the limit of a product, we obtain that$\lim_{{{x}\to{a}}}{\left({f{{\left({x}\right)}}}-{f{{\left({a}\right)}}}\right)}=\lim_{{{x}\to{a}}}\frac{{{f{{\left({x}\right)}}}-{f{{\left({a}\right)}}}}}{{{x}-{a}}}\lim_{{{x}\to{a}}}{\left({x}-{a}\right)}$, or $\lim_{{{x}\to{a}}}{\left({f{{\left({x}\right)}}}-{f{{\left({a}\right)}}}\right)}={f{'}}{\left({a}\right)}\cdot{\left({a}-{a}\right)}$.

Thus, $\lim_{{{x}\to{a}}}{\left({f{{\left({x}\right)}}}-{f{{\left({a}\right)}}}\right)}={0}$.

Now, $\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}=\lim_{{{x}\to{a}}}{\left({f{{\left({x}\right)}}}-{f{{\left({a}\right)}}}+{f{{\left({a}\right)}}}\right)}=$

$=\lim_{{{x}\to{a}}}{\left({f{{\left({x}\right)}}}-{f{{\left({a}\right)}}}\right)}+\lim_{{{x}\to{a}}}{\left({f{{\left({a}\right)}}}\right)}={0}+{f{{\left({a}\right)}}}={f{{\left({a}\right)}}}$.

Therefore, $\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}={f{{\left({a}\right)}}}$, which means that the function is continuous at ${a}$.

The converse is not true: if ${f{}}$ is continuous at ${a}$, it doesn't mean that ${f{}}$ is differentiable at ${a}$. Consider the function ${f{{\left({x}\right)}}}={\left|{x}\right|}$. It is continuous at ${x}={0}$, but, as was shown in example 4, it is not differentiable at $0$.