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Definition of Derivative

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There is one limit that is used very frequently in the applications of calculus and different sciences. This limit has the form lim⁡h→0f(x+h)−f(x)h\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 ff is a function that is denoted by f′(x){f{'}}{\left({x}\right)} and is calculated as f′(x)=lim⁡h→0f(x+h)−f(x)hf{'}{\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: dydx\frac{{{d}{y}}}{{{d}{x}}}, dfdx\frac{{{d}{f}}}{{{d}{x}}}, ddxf(x)\frac{{d}}{{{d}{x}}}{f{{\left({x}\right)}}}. Note that here dydx\frac{{{d}{y}}}{{{d}{x}}} is not a ratio, it is just a synonym for f′(x).{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{a}, we use the notation dydx∣x=a\frac{{{d}{y}}}{{{d}{x}}}{\mid}_{{{x}={a}}}.

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

Cauchy's notation: Dy{D}{y}, Df(x){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 yx′{y}'_{{x}}, f′x{f{'}}_{{x}}, Dxy{D}_{{x}}{y}, Dxf{D}_{{x}}{f{}}.

We will be mainly using Lagrange's notation.

Example 1. If f(x)=x2{f{{\left({x}\right)}}}={{x}}^{{2}}, find f′(x){f{'}}{\left({x}\right)} and f′(3){f'{{\left({3}\right)}}}.

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

Thus,

f′(x)=lim⁡h→0f(x+h)−f(x)h=lim⁡h→0(x+h)2−(x)2h=lim⁡h→0x2+2xh+h2−x2h={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→02xh+h2h=lim⁡h→0(2x+h)=2x=\lim_{{{h}\to{0}}}\frac{{{2}{x}{h}+{{h}}^{{2}}}}{{h}}=\lim_{{{h}\to{0}}}{\left({2}{x}+{h}\right)}={2}{x}.

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

Let's work another example.

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

f′(x)=lim⁡h→0x+h−xh=lim⁡h→0x+h−xhx+h+xx+h+x=lim⁡h→0x+h−xh(x+h+x)={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→01x+h+x=1x+0+x=12x=\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′{f{'}} is (0,∞){\left({0},\infty\right)}. As can be seen, it is smaller than the domain of f(x)=x{f{{\left({x}\right)}}}=\sqrt{{{x}}}.

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

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

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

We have that f′(x)=lim⁡h→0f(x+h)−f(x)h{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{h}={0.5}, we will obtain the following approximation: f′(x)≈f(x+0.5)−f(x)0.5=2(f(x+0.5)−f(x)){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′(1)≈2(f(1+0.5)−f(1))=2(f(1.5)−f(1))=2(3−2)=2{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′(1.5)≈2(f(1.5+0.5)−f(1.5))=2(f(2)−f(1.5))=2(7−3)=8{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{f{}} is differentiable at a{a} if f′(a){f{'}}{\left({a}\right)} exists.

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

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

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

Therefore, for x>0{x}>{0}, f′(x)=lim⁡h→0∣x+h∣−∣x∣h=lim⁡h→0x+h−xh=lim⁡h→01=1{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{f{}} is differentiable for any x>0{x}>{0}.

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

Therefore, for x<0{x}<{0}, f′(x)=lim⁡h→0∣x+h∣−∣x∣h=lim⁡h→0−(x+h)−(−x)h=lim⁡h→0−1=−1{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{f{}} is differentiable for any x<0{x}<{0}.

For x=0{x}={0}, we have to investigatef′(0)=lim⁡h→0∣0+h∣−∣0∣h=lim⁡h→0∣h∣h{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→0+∣h∣h=lim⁡h→0+hh=lim⁡h→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}

lim⁡h→0−∣h∣h=lim⁡h→0−−hh=lim⁡h→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→0+∣h∣h≠lim⁡h→0−∣h∣h\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→0∣h∣h\lim_{{{h}\to{0}}}\frac{{{\left|{h}\right|}}}{{h}} doesn't exist and thus f(x){f{{\left({x}\right)}}} is not differentiable at x=0.{x}={0}.

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

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

Proof. Since f{f{}} is differentiable at a{a}, we can say that f′(a)=lim⁡h→0f(a+h)−f(a)h{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{x}={a}+{h}, we have that h=x−a{h}={x}-{a}. As h→0{h}\to{0}, x→a{x}\to{a}.

Therefore, f′(a)=lim⁡x→af(x)−f(a)x−a{f{'}}{\left({a}\right)}=\lim_{{{x}\to{a}}}\frac{{{f{{\left({x}\right)}}}-{f{{\left({a}\right)}}}}}{{{x}-{a}}}.

Now, consider the identity f(x)−f(a)=f(x)−f(a){f{{\left({x}\right)}}}-{f{{\left({a}\right)}}}={f{{\left({x}\right)}}}-{f{{\left({a}\right)}}}, or, equivalently, f(x)−f(a)=f(x)−f(a)x−a(x−a){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→a(f(x)−f(a))=lim⁡x→a(f(x)−f(a)x−a(x−a))\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 thatlim⁡x→a(f(x)−f(a))=lim⁡x→af(x)−f(a)x−alim⁡x→a(x−a)\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→a(f(x)−f(a))=f′(a)⋅(a−a)\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→a(f(x)−f(a))=0\lim_{{{x}\to{a}}}{\left({f{{\left({x}\right)}}}-{f{{\left({a}\right)}}}\right)}={0}.

Now, lim⁡x→af(x)=lim⁡x→a(f(x)−f(a)+f(a))=\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→a(f(x)−f(a))+lim⁡x→a(f(a))=0+f(a)=f(a)=\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→af(x)=f(a)\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}={f{{\left({a}\right)}}}, which means that the function is continuous at a{a}.

The converse is not true: if f{f{}} is continuous at a{a}, it doesn't mean that f{f{}} is differentiable at a{a}. Consider the function f(x)=∣x∣{f{{\left({x}\right)}}}={\left|{x}\right|}. It is continuous at x=0{x}={0}, but, as was shown in example 4, it is not differentiable at 00.