6.2: Derivatives

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- 6.1: Best Linear Approximations
- 6.3: Mean Value Theorem

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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 $(6.2 .14)$ ,

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

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

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

$f^{-1}(f(a)+h)=a+k,$

$f(a)+h=f(a+k)$

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 $f^{\prime}(x)=n x^{n-1}$ .

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Proposition $\PageIndex{2}$

Exercise $\PageIndex{5}$

Proposition $\PageIndex{3}$

Exercise $\PageIndex{6}$

Proposition $\PageIndex{4}$

Exercise $\PageIndex{7}$

Proposition $\PageIndex{5}$

Exercise $\PageIndex{8}$

Proposition $\PageIndex{6}$

Proposition $\PageIndex{7}$

Example $\PageIndex{3}$

Exercise $\PageIndex{9}$

Definition

Definition

Example 6.2.1\PageIndex{1}

Example 6.2.2\PageIndex{2}

Exercise 6.2.1\PageIndex{1}

Exercise 6.2.2\PageIndex{2}

Exercise 6.2.3\PageIndex{3}

Exercise 6.2.4\PageIndex{4}

Proposition 6.2.1\PageIndex{1}

6.2.1 The Rules

Proposition 6.2.2\PageIndex{2}

Exercise 6.2.5\PageIndex{5}

Proposition 6.2.3\PageIndex{3}

Exercise 6.2.6\PageIndex{6}

Proposition 6.2.4\PageIndex{4}

Exercise 6.2.7\PageIndex{7}

Proposition 6.2.5\PageIndex{5}

Exercise 6.2.8\PageIndex{8}

Proposition 6.2.6\PageIndex{6}

Proposition 6.2.7\PageIndex{7}

Example 6.2.3\PageIndex{3}

Exercise 6.2.9\PageIndex{9}

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Example $\PageIndex{1}$

Example $\PageIndex{2}$

Exercise $\PageIndex{1}$

Exercise $\PageIndex{2}$

Exercise $\PageIndex{3}$

Exercise $\PageIndex{4}$

Proposition $\PageIndex{1}$

Proposition $\PageIndex{2}$

Exercise $\PageIndex{5}$

Proposition $\PageIndex{3}$

Exercise $\PageIndex{6}$

Proposition $\PageIndex{4}$

Exercise $\PageIndex{7}$

Proposition $\PageIndex{5}$

Exercise $\PageIndex{8}$

Proposition $\PageIndex{6}$

Proposition $\PageIndex{7}$

Example $\PageIndex{3}$

Exercise $\PageIndex{9}$