4.1: Definition and Basic Properties of the Derivative

Let \(G\) be an open subset of \(\mathbb{R}\) and consider a function \(f: G \rightarrow \mathbb{R}\). For every \(a \in G\), the function

\[\phi_{a}(x)=\frac{f(x)-f(a)}{x-a}\]

is defined on \(G \backslash\{a\}\). Since \(G\) is an open set, \(a\) is a limit point of \(G \backslash \{a\}\) (see Example 2.6.6). Thus, it is possible to discuss the limit

\[\lim _{x \rightarrow a} \phi_{a}(x)=\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}.\]

Definition \(\PageIndex{1}\): Differentiable and Derivative

Let \(G\) be an open subset of \(\mathbb{R}\) and let \(a \in G\). We say that the function \(f\) defined on \(G\) is differentiable at \(a\) if the limit

\[\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}\]

exists (as a real number). In this case, the limit is called the derivative of \(f\) at \(a\) denoted by \(f^{\prime}(a)\), and \(f\) is said to be differentiable at \(a\). Thus, if \(f\) is differentiable at \(a\), then

\[f^{\prime}(a)=\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}.\]

We say that \(f\) is differentiable on \(G\) if \(f\) is differentiable at every point \(a \in G\). In this case, the function \(f^{\prime}: G \rightarrow \mathbb{R}\) is called the derivative on \(f\) on \(G\).

Example \(\PageIndex{1}\)

1. Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be given by \(f(x)=x\) and let \(a \in \mathbb{R}\).
2. Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be given by \(f(x)=x^{2}\) and let \(a \in \mathbb{R}\).
3. Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be given by \(f(x)=|x|\) and let \(a = 0\).

Solution

1. Then

\[\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}=\lim _{x \rightarrow a} \frac{x-a}{x-a}=\lim _{x \rightarrow a} 1=1.\]

It follows that \(f\) is differentiable at \(a\) and \(f^{\prime}(a)=1\).

2. Then

\[\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}=\lim _{x \rightarrow a} \frac{x^{2}-a^{2}}{x-a}=\lim _{x \rightarrow a} \frac{(x-a)(x+a)}{x-a}=\lim _{x \rightarrow a}(x+a)=2 a .\]

Thus, \(f\) is differentiable at every \(a \in \mathbb{R}\) and \(f^{\prime}(a)=2 a\).

3. Then

\[\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0^{+}} \frac{|x|}{x}=\lim _{x \rightarrow 0^{+}} \frac{x}{x}=1 ,\]

and

\[\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0^{-}} \frac{|x|}{x}=\lim _{x \rightarrow 0^{-}} \frac{-x}{x}=-1 .\]

Therefore, \(\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}\) does not exists and, hence, \(f\) is not differentiable at \(0\).

Theorem \(\PageIndex{1}\)

Let \(G\) be an open subset of \(\mathbb{R}\) and let \(f\) be defined on \(G\). If \(f\) is differentiable at \(a \in G\), then \(f\) is continuous at this point.

Proof

We have the following identity for \(x \in G \backslash\{a\}\):

\[\begin{aligned}
f(x) &=f(x)-f(a)+f(a) \\
&=\frac{f(x)-f(a)}{x-a}(x-a)+f(a)
\end{aligned} .\]

Thus,

\[\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a}\left[\frac{f(x)-f(a)}{x-a}(x-a)+f(a)\right]=f^{\prime}(a) \cdot 0+f(a)=f(a) .\]

Therefore, \(f\) is continuous at \(a\) by Theorem 3.3.2. \(\square\)

Remark \(\PageIndex{2}\)

The converse of Theorem 4.1.1 is not true. For instance, the absolute value function \(f(x)=|x|\) is continuous at \(0\), but it is not differentiable at this point (as shown in the example above).

Theorem \(\PageIndex{3}\)

Let \(G\) be an open subset of \(\mathbb{R}\) and let \(f, g: G \rightarrow \mathbb{R}\). Suppose both \(f\) and \(g\) are differentiable at \(a \in G\). Then the following hold.

1. The function \(f+g\) is differnetiable at \(a\) and \[(f+g)^{\prime}(a)=f^{\prime}(a)+g^{\prime}(a) .\]
2. For a constant \(c\), the function \(cf\) is differentiable at \(a\) and \[(c f)^{\prime}(a)=c f^{\prime}(a) .\]
3. The function \(fg\) is differentiable at \(a\) and \[(f g)^{\prime}(a)=f^{\prime}(a) g(a)+f(a) g^{\prime}(a) .\]
4. Suppose additionally that \(g(a) \neq 0\). Then the function \(\frac{f}{g}\) is differentiable at \(a\) and \[\left(\frac{f}{g}\right)^{\prime}(a)=\frac{f^{\prime}(a) g(a)-f(a) g^{\prime}(a)}{(g(a))^{2}}\].

Proof

The proofs for (a) and (b) are straightforward and we leave them as exsercises. Let us prove (c). For every \(x \in G \backslash\{a\}\), we can write

\[\begin{aligned}
\frac{(f g)(x)-(f g)(a)}{x-a} &=\frac{f(x) g(x)-f(a) g(x)+f(a) g(x)-f(a) g(a)}{x-a} \\
&=\frac{(f(x)-f(a)) g(x)}{x-a}+\frac{f(a)(g(x)-g(a))}{x-a}
\end{aligned} .\]

By Theorem 4.1.1, the function \(g\) is continuous at \(a\) and, hence

\[\lim _{x \rightarrow a} g(x)=g(a) .\]

Thus,

\[\lim _{x \rightarrow a} \frac{(f g)(x)-(f g)(a)}{x-a}=f^{\prime}(a) g(a)+f(a) g^{\prime}(a) .\]

This implies (c).

Next we show (d). Since \(g(a) \neq 0\), by (4.1), there exists an open interval \(I\) containing \(a\) such that \(g(x) \neq 0\) for all \(x \in I\). Let \(h = \frac{f}{g}\). Then \(h\) is defined on \(I\). Moreover,

\[\begin{aligned}
\frac{h(x)-h(a)}{x-a} &=\frac{\frac{f(x)}{g(x)}-\frac{f(a)}{g(x)}+\frac{f(a)}{g(x)}-\frac{f(a)}{g(a)}}{x-a} \\
&=\frac{\frac{1}{g(x)}(f(x)-f(a))+\frac{f(a)}{g(x) g(a)}(g(a)-g(x))}{x-a} \\
&=\frac{1}{g(x) g(a)}\left[g(a) \frac{f(x)-f(a)}{x-a}-f(a) \frac{g(x)-g(a)}{x-a}\right]
\end{aligned} .\]

Taking the limit as \(x \rightarrow a\), we obtain (d). The proof is now complete. \(\square\)

Example \(\PageIndex{2}\)

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be given by \(f(x) = x^2\) and let \(a \in \mathbb{R}\).

Solution

Using Example 4.1.1(a) and Theorem 4.1.3(c) we can provide an alternative derivation of a formula for \(f^{\prime}(a)\). Indeed, let \(g: \mathbb{R} \rightarrow \mathbb{R}\) be given by \(g(x) = x\). Then \(f=g \cdot g\) so

\[f^{\prime}(a)=(g g)^{\prime}(a)=g^{\prime}(a) g(a)+g(a) g^{\prime}(a)=2 g^{\prime}(a) g(a)=2 a .\]

Proceeding by induction, we can obtain the derivative of \(g: \mathbb{R} \rightarrow \mathbb{R}\) given by \(g(x)=x^{n}\) for \(n \in \mathbb{N}\) as \(g^{\prime}(a)=n x^{n-1}\). Furthermore, using this and Theorem 4.1.3(a)(b) we obtain the familiar formula for the derivative of a polynomial \(p(x)=a_{n} x^{n}+\cdots+a_{1} x+a_{0}\) as \(p^{\prime}(x)=n a_{n} x^{n-1}+\cdots+2 a_{2} x+a_{1}\).

The following lemma is very convenient for studying the differentiability of the composition of functions.

Lemma \(\PageIndex{4}\)

Let \(G\) be an open subset of \(\mathbb{R}\) and let \(f: G \rightarrow \mathbb{R}\). Suppose \(f\) is differentiable at \(a\). Then there exists a function \(u: G \rightarrow \mathbb{R}\) satisfying

\[f(x)-f(a)=\left[f^{\prime}(a)+u(x)\right](x-a) \text { for all } x \in G\]

and \(\lim _{x \rightarrow a} u(x)=0\).

Proof

Define

\[u(x)=\left\{\begin{array}{ll}
\frac{f(x)-f(a)}{x-a}-f^{\prime}(a), & x \in G \backslash\{a\} \\
0, & x=a
\end{array}\right .\]

Since \(f\) is differentiable at \(a\), we have

\[\lim _{x \rightarrow a} u(x)=\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}-f^{\prime}(a)=f^{\prime}(a)-f^{\prime}(a)=0 .\]

Therefore, the function \(u\) satisfies the conditions of the lemma. \(\square\)

Theorem \(\PageIndex{5}\) - Chain rule.

Let \(f: G_{1} \rightarrow \mathbb{R}\) and let \(g: G_{2} \rightarrow \mathbb{R}\), where \(G_{1}\) and \(G_{2}\) are two open subsets of \(\mathbb{R}\) with \(f\left(G_{1}\right) \subset G_{2}\). Suppose \(f\) is differentiable at \(a\) and \(g\) is differentiable at \(f(a)\). Then the function \(g \circ f\) is differentiable at \(a\) and

\[(g \circ f)^{\prime}(a)=g^{\prime}(f(a)) f^{\prime}(a) .\]

Proof

Since \(f\) is differentiable at \(a\), by Lemma 4.1.4, there exists a function \(u\) defined on \(G_{1}\) with

\[f(x)-f(a)=\left[f^{\prime}(a)+u(x)\right](x-a) \text { for all } x \in G_{1} ,\]

and \(\lim _{x \rightarrow a} u(x)=0\).

Similarly, since \(g\) is differentiable at \(f(a)\), there exists a function \(v\) defined on \(G_{2}\) with

\[g(t)-g(f(a))=\left[g^{\prime}(f(a))+v(t)\right][t-f(a)] \text { for all } t \in G_{2} ,\]

and \(\lim _{t \rightarrow f(a)} v(t)=0\).

Applying (4.2) for \(t = f(x)\), we have

\[g(f(x))-g(f(a))=\left[g^{\prime}(f(a))+v(f(x))\right][f(x)-f(a)] .\]

Thus,

\[g(f(x))-g(f(a))=\left[g^{\prime}(f(a))+v(f(x))\right]\left[f^{\prime}(a)+u(x)\right](x-a) \text { for all } x \in G_{1} .\]

This implies

\[\frac{g(f(x))-g(f(a))}{x-a}=\left[g^{\prime}(f(a))+v(f(x))\right]\left[f^{\prime}(a)+u(x)\right] \text { for all } x \in G_{1} \backslash\{a\} .\]

By the continuity of \(f\) at \(a\) and the property of \(v\), we have \(\lim _{x \rightarrow a} v(f(x))=0\) and, hence,

\[\lim _{x \rightarrow a} \frac{g(f(x))-g(f(a))}{x-a}=g^{\prime}(f(a)) f^{\prime}(a)\].

The proof is now complete. \(\square\)

Example \(\PageIndex{3}\)

Consider the function \(h: \mathbb{R} \rightarrow \mathbb{R}\) given by \(h(x)=\left(3 x^{4}+x+7\right)^{15}\).

Solution

Since \(h(x)\) is a polynomial we could in principle compute \(h^{\prime}(x)\) by expanding the power and using Example 4.1.2. However, Theorem 4.1.5 provides a shorter way. Define \(f, g: \mathbb{R} \rightarrow \mathbb{R}\) by \(f(x)=3 x^{4}+x+7\) and \(g(x)=x^{15}\). Then \(h=g \circ f\). Given \(a \in \mathbb{R}\), it follows from Theorem 4.1.5 that

\[(g \circ f)^{\prime}(a)=g^{\prime}(f(a)) f^{\prime}(a)=15\left(3 a^{4}+a+7\right)^{14}\left(12 a^{3}+1\right) .\]

Example \(\PageIndex{4}\)

By iterating the Chain Rule, we can extended the result to the composition of more than two functions in a straightforward way. For example, given functions \(f: G_{1} \rightarrow \mathbb{R}\), \(g: G_{2} \rightarrow \mathbb{R}\), and \(h: G_{3} \rightarrow \mathbb{R}\) such that \(f\left(G_{1}\right) \subset G_{2}\), \(g\left(G_{2}\right) \subset G_{3}\), \(f\) is differentiable at \(a\), \(g\) is differentiable at \(f(a)\), and \(h\) is differentiable at \(g(f(a))\), we obtain that \(h \circ g \circ f\) is differentiable at \(a\) and \((h \circ g \circ f)^{\prime}(a)=h^{\prime}(g(f(a))) g^{\prime}(f(a)) f^{\prime}(a)\)

Solution

Add text here.

Definition \(\PageIndex{2}\)

Let \(G\) be an open set and let \(f: G \rightarrow \mathbb{R}\) be a differentiable function. If the function \(f^{\prime}: G \rightarrow \mathbb{R}\) is also differentiable, we say that \(f\) is twice differentiable (on \(G\)). The second derivative of \(f\) is denoted \(f^{\prime \prime}\) or \(f^{(2)}\). Thus \(f^{\prime \prime}=\left(f^{\prime}\right)^{\prime}\). Similarly, we say that \(f\) is three times differentiable if \(f^{(2)}\) is differentiable, and \(\left(f^{(2)}\right)^{\prime}\) is called the third derivative of \(f\) and is denoted by \(f^{\prime \prime \prime}\) or \(f^{(3)}\). We can define in this way \(n\) times differentiability and the \(nth\) derivative of \(f\) for any positive integer \(n\). As a convention, \(f^{(0)}=f\).

Definition \(\PageIndex{3}\)

Let \(I\) be an open interval in \(\mathbb{R}\) and let \(f: I \rightarrow \mathbb{R}\). The function \(f\) is said to be continuously differentiable if \(f\) is differentiable on \(I\) and \(f^{\prime}\) is continuous on \(I\). We denote by \(C^{1}(I)\) the set of all continuously differentiable functions on \(I\). If \(f\) is \(n\) times differentiable on \(I\) and the \(nth\) derivative is continuous, then \(f\) is called \(n\) times continuously differentiable. We denote by \(C^{n}(I)\) the set of all \(n\) times continuously differentiable functions on \(I\).

Exercise \(\PageIndex{1}\)

Prove parts (a) and (b) of Theorem 4.1.3.

Answer

Add texts here. Do not delete this text first.

Exercise \(\PageIndex{2}\)

Compute the following derivatives directly from the definition. That is, do not use Theorem 4.1.3, but rather compute the appropriate limit directly (see Example 4.1.1).

1. \(f(x)=m x+b \text { where } m, b \in \mathbb{R}\).
2. \(f(x) = \frac{1}{x}\) (here assume \(x \neq 0\).
3. \(f(x)=\sqrt{x}\) (here assume \(x > 0\))

Answer

Add texts here. Do not delete this text first.

Exercise \(\PageIndex{3}\)

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be given by

\[f(x)=\left\{\begin{array}{ll}
x^{2}, & \text { if } x>0 \text {;}\\
0, & \text { if } x \leq 0 \text {.}
\end{array}\right.\]

1. Prove that \(f\) is differentiable at \(0\). Find \(f^{\prime}(x)\) for all \(x \in \mathbb{R}\).
2. Is \(f^{\prime}\) continuous? Is \(f^{\prime}\) differentiable?

Answer

Add texts here. Do not delete this text first.

Exercise \(\PageIndex{4}\)

Let

\[f(x)=\left\{\begin{array}{ll}
x^{\alpha}, & \text { if } x>0 \text {;} \\
0, & \text { if } x \leq 0 \text {.}
\end{array}\right.\]

1. Determine the values of \(\alpha\) for which \(f\) is continuous on \(\mathbb{R}\).
2. Determine the values of \(\alpha\) for which \(f\) is differentiable on \(\mathbb{R}\). In this case, find \(f^{\prime}\).

Answer

Add texts here. Do not delete this text first.

Exercise \(\PageIndex{5}\)

Use Theorems 4.1.3 and 4.1.5 to compute the derivatives of the following functions at the indicated points (see also Example 4.1.4). (Assume known that the function \(\sin x\) is differentiable at all points and that its derivative is \(\cos x\).)

1. \(f(x)=\frac{3 x^{4}+7 x}{2 x^{2}+3} \text { at } a=-1\).
2. \(f(x)=\sin ^{5}\left(3 x^{2}+\frac{\pi}{2} x\right) \text { at } a=\frac{\pi}{8}\)

Answer

Add texts here. Do not delete this text first.

Exercise \(\PageIndex{6}\)

Determine the values of \(x\) at which each function is differentiable.

1. \(f(x)=\left\{\begin{array}{ll}
   x \sin \frac{1}{x}, & \text { if } x \neq 0 \text {;} \\
   0, & \text { if } x=0 \text {.}
   \end{array}\right.\)
2. \(f(x)=\left\{\begin{array}{ll}
   x^{2} \sin \frac{1}{x}, & \text { if } x \neq 0 \text {;} \\
   0, & \text { if } x=0 \text {;}
   \end{array}\right.\)

Answer

Add texts here. Do not delete this text first.

Exercise \(\PageIndex{7}\)

Determine if each of the following functions is differentiale at \(0\). Justify your answer.

1. f(x)=\left\{\begin{array}{ll}
   x^{2}, & \text { if } x \in \mathbb{Q} \text {;} \\
   x^{3}, & \text { if } x \notin \mathbb{Q} \text {.}
   \end{array}\right.\)
2. \(f(x)=[x] \sin ^{2}(\pi x)\).
3. \(f(x)=\cos (\sqrt{|x|})\).
4. \(f(x)=x|x|\).

Answer

Add texts here. Do not delete this text first.

Exercise \(\PageIndex{8}\)

Let \(f, g\) be differentiable at \(a\). Find the following limits:

1. \(\lim _{x \rightarrow a} \frac{x f(a)-a f(x)}{x-a}\).
2. \(\lim _{x \rightarrow a} \frac{f(x) g(a)-f(a) g(x)}{x-a}\).

Answer

Add texts here. Do not delete this text first.

Exercise \(\PageIndex{9}\)

Let \(G\) be an open subset of \(\mathbb{R}\) and \(a \in G\). Prove that if \(f: G \rightarrow \mathbb{R}\) is Lipschitz continuous, then \(g(x)=(f(x)-f(a))^{2}\) is differentiable at \(a\).

Answer

Add texts here. Do not delete this text first.

Exercise \(\PageIndex{10}\)

Let \(f\) be differnetiable at \(a\) and \(f(a)>0\). Find the following limit:

\[\lim _{n \rightarrow \infty}\left(\frac{f\left(a+\frac{1}{n}\right)}{f(a)}\right)^{n}\].

Answer

Add texts here. Do not delete this text first.

Exercise \(\PageIndex{11}\)

Consider the function

\[f(x)=\left\{\begin{array}{ll}
x^{2} \sin \frac{1}{x}+c x, & \text { if } x \neq 0 \text {;}\\
0, & \text { if } x=0 \text {,}
\end{array}\right.\]

where \(0<c<1\).

1. Prove that the function is differentiable on \(\mathbb{R}\).
2. Prove that for every \(\alpha > 0\), the function \(f^{\prime}\) changes its sign on \((-\alpha, \alpha)\).

Answer

Add texts here. Do not delete this text first.

Exercise \(\PageIndex{12}\)

Let \(f\) be differentiable at \(x_{0} \in(a, b)\) and let \(c\) be a constant. Prove that

1. \(\lim _{n \rightarrow \infty} n\left[f\left(x_{0}+\frac{1}{n}\right)-f\left(x_{0}\right)\right]=f^{\prime}\left(x_{0}\right)\).
2. \(\lim _{h \rightarrow 0} \frac{f\left(x_{0}+c h\right)-f\left(x_{0}\right)}{h}=c f^{\prime}\left(x_{0}\right)\).

Answer

Add texts here. Do not delete this text first.

Exercise \(\PageIndex{13}\)

Let \(f\) be differnetiable at \(x_{0} \in(a, b)\) and let \(c\) be a constant. Find the limit

\[\lim _{h \rightarrow 0} \frac{f\left(x_{0}+c h\right)-f\left(x_{0}-c h\right)}{h}.\]

Answer

Add texts here. Do not delete this text first.

Exercise \(\PageIndex{14}\)

Prove that \(f: \mathbb{R} \rightarrow \mathbb{R}\), given by \(f(x)=|x|^{3}, i\), is in \(C^{2}(\mathbb{R})\) but not in \(C^{3}(\mathbb{R})\) (refer to Definition 4.1.3). (Hint: the key issue is differnetiability at 0.)

Answer

Add texts here. Do not delete this text first.