6.2: Derivatives

Definition

Suppose \(D \subset \mathbb{R}, f: D \rightarrow \mathbb{R}, a\) is an interior point of \(D,\) and \(f\) is differentiable at \(a .\) We call

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

the derivative of \(f\) at \(a,\) which we denote \(f^{\prime}(a)\).

Note that if \(f\) is differentiable at \(a,\) then

\[\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}.\]

Definition

Suppose \(D \subset \mathbb{R}, f: D \rightarrow \mathbb{R},\) and \(E\) is the set of interior points of \(D\) at which \(f\) is differentiable. We call the function \(f^{\prime}: E \rightarrow \mathbb{R}\) defined by

\[f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\]

the derivative of \(f\).

Example \(\PageIndex{1}\)

Let \(n \in \mathbb{Z}^{+}\) and define \(f: \mathbb{R} \rightarrow \mathbb{R}\) by \(f(x)=x^{n} .\) Then

\[\begin{aligned} f^{\prime}(x) &=\lim _{h \rightarrow 0} \frac{(x+h)^{n}-x^{n}}{h} \\ &=\lim _{h \rightarrow 0} \frac{x^{n}+n x^{n-1} h+\sum_{k=2}^{n}\left(\begin{array}{l}{n} \\ {k}\end{array}\right) x^{n-k} h^{k}-x^{n}}{h} \\ &=\lim _{h \rightarrow 0}\left(n x^{n-1}+\sum_{k=2}^{n}\left(\begin{array}{l}{n} \\ {k}\end{array}\right) x^{n-k} h^{k-1}\right) \\ &=n x^{n-1}.\end{aligned}\]

Example \(\PageIndex{2}\)

Define \(f: \mathbb{R} \rightarrow \mathbb{R}\) by \(f(x)=|x| .\) Then

\[\frac{f(0+h)-f(0)}{h}=\frac{|h|}{h}=\left\{\begin{aligned} 1, & \text { if } h>0, \\-1, & \text { if } h<0. \end{aligned}\right.\]

Hence

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

and

\[\lim _{h \rightarrow 0+} \frac{f(0+h)-f(0)}{h}=1.\]

Thus \(f\) is not differentiable at \(0 .\)

Exercise \(\PageIndex{1}\)

Show that if \(c \in \mathbb{R}\) and \(f(x)=c\) for all \(x \in \mathbb{R},\) then \(f^{\prime}(x)=0\) for all \(x \in \mathbb{R}\).

Exercise \(\PageIndex{2}\)

Define \(f:[0,+\infty) \rightarrow[0,+\infty)\) by \(f(x)=\sqrt{x} .\) Show that \(f^{\prime}:(0,+\infty) \rightarrow(0,+\infty)\) is given by

\[f^{\prime}(x)=\frac{1}{2 \sqrt{x}}.\]

Exercise \(\PageIndex{3}\)

Define \(f: \mathbb{R} \rightarrow \mathbb{R}\) by

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

Is \(f\) differentiable at \(0 ?\)

Exercise \(\PageIndex{4}\)

Define \(f: \mathbb{R} \rightarrow \mathbb{R}\) by

\[f(x)=\left\{\begin{array}{ll}{x^{2},} & {\text { if } x<0,} \\ {x^{3},} & {\text { if } x \geq 0.}\end{array}\right.\]

Is \(f\) differentiable at \(0 ?\)

Proposition \(\PageIndex{1}\)

If \(f\) is differentiable at \(a\), then \(f\) is continuous at \(a\).

Proof

If \(f\) is differentiable at \(a,\) then

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

Hence \(\lim _{x \rightarrow a} f(x)=f(a),\) and so \(f\) is continuous at \(a\). \(\quad\) Q.E.D.

6.2.1 The Rules

Proposition \(\PageIndex{2}\)

Suppose \(f\) is differentiable at \(a\) and \(\alpha \in \mathbb{R} .\) Then \(\alpha f\) is differentiable at \(a\) and \((\alpha f)^{\prime}(a)=\alpha f^{\prime}(a)\).

Exercise \(\PageIndex{5}\)

Prove the previous proposition.

Proposition \(\PageIndex{3}\)

Suppose \(f\) and \(g\) are both differentiable at \(a .\) Then \(f+g\) is differentiable at \(a\) and \((f+g)^{\prime}(a)=f^{\prime}(a)+g^{\prime}(a)\).

Exercise \(\PageIndex{6}\)

Prove the previous proposition.

Proposition \(\PageIndex{4}\)

(Product rule).

Suppose \(f\) and \(g\) are both differentiable at a. Then \(f g\) is differentiable at \(a\) and

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

Proof

We have

\[\begin{aligned}(f g)^{\prime}(a) &=\lim _{h \rightarrow 0} \frac{f(a+h) g(a+h)-f(a) g(a)}{h} \\ &=\lim _{h \rightarrow 0} \frac{f(a+h) g(a+h)-f(a) g(a+h)+f(a) g(a+h)-f(a) g(a)}{h} \\ &=\lim _{h \rightarrow 0}\left(g(a+h) \frac{f(a+h)-f(a)}{h}+f(a) \frac{g(a+h)-g(a)}{h}\right) \\ &=g(a) f^{\prime}(a)+f(a) g^{\prime}(a). \end{aligned}\]

where we know \(\lim _{h \rightarrow 0} g(a+h)=g(a)\) by the continuity of \(g\) at \(a,\) which in turn follows from the assumption that \(g\) is differentiable at \(a .\) \(\quad\) Q.E.D.

Exercise \(\PageIndex{7}\)

Given \(n \in \mathbb{Z}^{+}\) and \(f(x)=x^{n},\) use induction and the product rule to show that \(f^{\prime}(x)=n x^{n-1}\).

Proposition \(\PageIndex{5}\)

(Quotient rule).

Suppose \(D \subset \mathbb{R}, f: D \rightarrow \mathbb{R}, g: D \rightarrow \mathbb{R}\), \(a\) is in the interior of \(D\), and \(g(x) \neq 0\) for all \(x \in D .\) If \(f\) and \(g\) are both differentiable at \(a\), then \(\frac{f}{g}\) is differentiable at \(a\) and

\[\left(\frac{f}{g}\right)^{\prime}(a)=\frac{g(a) f^{\prime}(a)-f(a) g^{\prime}(a)}{(g(a))^{2}}.\]

Proof

\[\begin{aligned}\left(\frac{f}{g}\right)^{\prime}(a) &=\lim _{h \rightarrow 0} \frac{\frac{f(a+h)}{g(a+h)}-\frac{f(a)}{g(a)}}{h} \\ &=\lim _{h \rightarrow 0} \frac{f(a+h) g(a)-f(a) g(a+h)}{h g(a+h) g(a)} \\ &=\lim _{h \rightarrow 0} \frac{f(a+h) g(a)-f(a) g(a)+f(a) g(a)-f(a) g(a+h)}{h g(a+h) g(a)} \\ &=\lim _{h \rightarrow 0} \frac{g(a)\frac{f(a+h)-f(a)}{h}-f(a) \frac{g(a+h)-g(a)}{h}}{g(a+h)g(a)} \\ &=\frac{g(a) f^{\prime}(a)-f(a) g^{\prime}(a)}{(g(a))^{2}}; \end{aligned}\]

where we know \(\lim _{h \rightarrow 0} g(a+h)=g(a)\) by the continuity of \(g\) at \(a,\) which in turn follows from the assumption that \(g\) is differentiable at \(a .\) \(\quad\) Q.E.D.

Exercise \(\PageIndex{8}\)

Show that for any integer \(n \neq 0,\) if \(f(x)=x^{n},\) then \(f^{\prime}(x)=n x^{n-1}\).

Proposition \(\PageIndex{6}\)

(Chain rule).

Suppose \(D \subset \mathbb{R}, E \subset \mathbb{R}, g: D \rightarrow \mathbb{R}, f: E \rightarrow\mathbb{R}, g(D) \subset E, g\) is differentiable at \(a,\) and \(f\) is differentiable at \(g(a) .\) Then \(f \circ g\) is differentiable at \(a\) and

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

Proof

Since \(a\) is an interior point of \(D\) and \(g(a)\) is an interior point of \(E,\) we may choose \(\delta>0\) so that \((a-\delta, a+\delta) \subset D\) and \(\epsilon>0\) so that \((g(a)-\epsilon, g(a)+\epsilon) \subset E\). Define \(\varphi:(-\delta, \delta) \rightarrow \mathbb{R}\) by

\[\varphi(h)=\left\{\begin{array}{ll}{\frac{g(a+h)-g(a)-g^{\prime}(a) h}{h},} & {\text { if } h \neq 0,} \\ {0,} & {\text { if } h=0,}\end{array}\right.\]

and \(\psi:(-\epsilon, \epsilon) \rightarrow \mathbb{R}\) by

\[\psi(h)=\left\{\begin{array}{ll}{\frac{f(g(a)+h)-f(g(a))-f^{\prime}(g(a)) h}{h},} & {\text { if } h \neq 0,} \\ {0,} & {\text { if } h=0.}\end{array}\right.\]

The assumption that \(g\) is differentiable at \(a\) implies that \(\varphi\) is continuous at 0 and the assumption that \(f\) is differentiable at \(g(a)\) implies that \(\psi\) is continuous at \(0 .\) Moreover, note that

\[g(a+h)=h \varphi(h)+g^{\prime}(a) h+g(a)\]

for \(h \in(-\delta, \delta)\) and

\[f(g(a)+h)=h \psi(h)+f^{\prime}(g(a)) h+f(g(a))\]

for \(h \in(-\epsilon, \epsilon) .\) From \((6.2 .12)\) we have

\[f(g(a+h))=f\left(h \varphi(h)+g^{\prime}(a) h+g(a)\right)\]

for \(h \in(-\delta, \delta) .\) Now

\[\lim _{h \rightarrow 0}\left(h \varphi(h)+g^{\prime}(a) h\right)=0,\]

so we may choose \(\gamma>0\) so that \(\gamma \leq \delta\) and

\[\left|h \varphi(h)+g^{\prime}(a) h\right|<\epsilon\]

whenever \(h \in(-\gamma, \gamma) .\) Thus, using \((6.2 .13)\) and \((6.2 .14)\),

\[f(g(a+h))=\left(h \varphi(h)+g^{\prime}(a) h\right) \psi\left(h \varphi(h)+g^{\prime}(a) h\right)+f^{\prime}(g(a))\left(h \varphi(h)+g^{\prime}(a) h\right)+f(g(a)),\]

so

\[\begin{aligned} f(g(a+h))-f(g(a))=&\left(h \varphi(h)+g^{\prime}(a) h\right) \psi\left(h \varphi(h)+g^{\prime}(a) h\right) \\ &+f^{\prime}(g(a))\left(h \varphi(h)+g^{\prime}(a) h\right) \\=& h \varphi(h) \psi\left(h \varphi(h)+g^{\prime}(a) h\right) \\ &+h g^{\prime}(a) \psi\left(h \varphi(h)+g^{\prime}(a) h\right) \\ &+f^{\prime}(g(a)) \varphi(h) h+f^{\prime}(g(a)) g^{\prime}(a) h. \end{aligned}\]

Hence

\[\begin{aligned} \frac{f(g(a+h))-f(g(a))}{h}=f^{\prime}(g(a)) g^{\prime}(a)+\varphi(h) \psi\left(h \varphi(h)+g^{\prime}(a) h\right) \\+g^{\prime}(a) \psi\left(h \varphi(h)+g^{\prime}(a) h\right)+f^{\prime}(g(a)) \varphi(h). \end{aligned}\]

Now

\[\lim _{h \rightarrow 0} \varphi(h)=0,\]

\[\lim _{h \rightarrow 0}\left(h \varphi(h)+g^{\prime}(a) h\right)=0,\]

and, since \(\varphi\) and \(\psi\) are continuous at 0,

\[\lim _{h \rightarrow 0} \psi\left(h \varphi(h)+g^{\prime}(a) h\right)=0.\]

Thus

\[\lim _{h \rightarrow 0} \frac{f(g(a+h))-f(g(a))}{h}=f^{\prime}(g(a)) g^{\prime}(a).\]

Q.E.D.

Proposition \(\PageIndex{7}\)

Suppose \(D \subset \mathbb{R}, f: D \rightarrow \mathbb{R}\) is one-to-one, \(a\) is in the interior of \(D, f(a)\) is in the interior of \(f(D), f^{-1}\) is continuous at \(f(a),\) and \(f\) is differentiable at \(a\) with \(f^{\prime}(a) \neq 0 .\) Then \(f^{-1}\) is differentiable at \(f(a)\) and

\[\left(f^{-1}\right)^{\prime}(f(a))=\frac{1}{f^{\prime}(a)}.\]

Proof

Choose \(\delta>0\) so that \((f(a)-\delta, f(a)+\delta) \subset f(D) .\) For \(h \in(-\delta, \delta),\) let

\[k=f^{-1}(f(a)+h)-a.\]

Then

\[f^{-1}(f(a)+h)=a+k,\]

so

\[f(a)+h=f(a+k)\]

and

\[h=f(a+k)-f(a).\]

Hence

\[\frac{f^{-1}(f(a)+h)-f^{-1}(f(a))}{h}=\frac{a+k-a}{f(a+k)-f(a)}=\frac{1}{\frac{f(a+k)-f(a)}{k}}.\]

Now if \(\left.h \rightarrow 0, \text { then } k \rightarrow 0 \text { (since } f^{-1} \text { is continuous at } f(a)\right),\) and so

\[\lim_{h \rightarrow 0} \frac{f^{-1}(f(a)+h)-f^{-1}(f(a))}{h}=\lim_{k \rightarrow 0} \frac{1}{\frac{f(a+k)-f(a)}{k}}=\frac{1}{f^{\prime}(a)}.\]

Q.E.D.

Example \(\PageIndex{3}\)

For \(n \in Z^{+},\) define \(f:[0,+\infty) \rightarrow \mathbb{R}\) by \(f(x)=\sqrt[n]{x} .\) Then \(f\) is the inverse of \(g:[0,+\infty) \rightarrow \mathbb{R}\) defined by \(g(x)=x^{n} .\) Thus, for any \(x \in(0,+\infty)\),

\[f^{\prime}(x)=\frac{1}{g^{\prime}(f(x))}=\frac{1}{n(\sqrt[n]{x})^{n-1}}=\frac{1}{n} x^{\frac{1}{n}-1}.\]

Exercise \(\PageIndex{9}\)

Let \(n \neq 0\) be a rational number and let \(f(x)=x^{n} .\) Show that \(f^{\prime}(x)=n x^{n-1}\).