Definition of Derivative

Related calculator: Online Derivative Calculator with Steps

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 the Leibniz 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 the Lagrange's and Cauchy's notations, if we need to explicitly state with respect to what variable we to 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.