11.1 Derivatives of Some Special Functions

11.1   Theorem (Derivative of power functions.) Let $r\in\mbox{{\bf Q}}$ and let
$f(x)=x^r$. Here

$${\rm domain} (f)=\cases{ \mbox{{\bf R}}&if $r\in\mbox{{\bf Z}... ...}}}^{+}$}&if $r\in\mbox{{\bf Q}}^-\setminus\mbox{{\bf Z}}$.\cr}$$

Let $a$ be an interior point of domain$(f)$. Then $f$ is differentiable at $a$, and

$$f^\prime (a)=ra^{r-1}.$$

If $r=0$ and $a=0$ we interpret $ra^{r-1}$ to be $0$.

Proof: First consider the case $a\neq 0$. For all $x$ in domain $(f)\setminus\{a\}$ we have

$${{f(x)-f(a)}\over {x-a}}={{x^r-a^r}\over {x-a}}={{a^r\Big( ({... ...=a^{r-1} {{\Big( ({x\over a})^r-1\Big)}\over {({x\over a}-1)}}.$$

Let $\{x_n\}$ be a generic sequence in domain $(f)\setminus\{a\}$ such that $\{x_n\}\to a$. Let $y_n=\displaystyle { {{x_n}\over a}}$. Then $\{y_n\}\to 1$ and hence by theorem 7.10 we have $${ \Big\{ {{y_n^r-1}\over {y_n-1}}\Big\}\to r}$$ and hence

$$\Bigg\{ {{f(x_n)-f(a)}\over {x_n-a}}\Bigg\}=\Bigg\{ {{y_n^r-1}\over {y_n-1}}\Bigg\}\cdot a^{r-1}\to ra^{r-1}.$$

This proves the theorem in the case $a\neq 0$. If $a=0$ then $r\in\mbox{{\bf Z}}_{\geq 0}$ (since for other values of $r$, $0$ is not an interior point of domain$(f)$). In this case

$${{f(x)-f(0)}\over{x-0}}={{x^r-0^r}\over x}=\cases{ 0 &if $r=0$ ({\rm remember} $0^0=1$).\cr x^{r-1} &if $r\neq 0$.\cr}$$

Hence

$$f^\prime (0)=\lim_{x\to 0} {{f(x)-f(0)}\over x}=\cases{ 0 &if $r=0$,\cr 1 &if $r=1$,\cr 0 &if $r>1$.\cr}$$

Thus in all cases the formula $f^\prime (x)=rx^{r-1}$ holds. $\diamondsuit$

11.2   Corollary (Of the proof of theorem 11.1) For all $r\in\mbox{{\bf Q}}$,

$$\lim_{x\to 1}{{x^r-1}\over {x-1}}=r.$$

11.3   Theorem (Derivatives of $\sin$ and $\cos$.) Let $r\in\mbox{{\bf R}}$ and let
$f(x)=\sin (rx),\; g(x)=\cos (rx)$ for all $x \in \mbox{{\bf R}}$. Then $f$ and $g$ are differentiable on $\mbox{{\bf R}}$, and for all $x \in \mbox{{\bf R}}$

$$f^\prime (x) = r\cos (rx),$$ (11.4)

$$g^\prime (x) = -r\sin (rx).$$ (11.5)

Proof: If $r=0$ the result is clear, so we assume $r \neq 0$. For all $x \in \mbox{{\bf R}}$ and all $t\in\mbox{{\bf R}}\setminus\{x\}$, we have

$$\begin{eqnarray*} {{\sin (rt)-\sin (rx)}\over {t-x}} &=& {{2\cos \Big( {{r(t+x)}... ...g( {{r(t-x)}\over 2}\Big)}\over {\Big( {{r(t-x)}\over 2}\Big)}}. \end{eqnarray*}$$

(Here I've used an identity from theorem 9.21.) Let $\{x_n\}$ be a generic sequence in $\mbox{{\bf R}}\setminus\{x\}$ such that $\{x_n\}\to x$. Let $y_n=\displaystyle {{{r(x_n+x)}\over 2}}$ and let $${z_n={{r(x_n-x)}\over 2}}$$. Then $\{y_n\}\to rx$ so by lemma 9.34 we have $\{\cos (y_n)\}\to\cos (rx)$. Also $\{z_n\} \to 0$, and $z_n \in \mbox{{\bf R}}\setminus \{0\}$ for all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$, so by (9.38), $${\Big\{{{\sin (z_n)}\over {z_n}}\Big\}\to 1}$$. Hence

$$\Big\{ {{\sin (rx_n)-\sin (rx)}\over {x_n-x}}\Big\}=\Big\{ r\cos (y_n)\cdot {{\sin (z_n)}\over {z_n}}\Big\}\to r\cos (rx),$$

and this proves formula (11.4). $\diamondsuit$

The proof of (11.5) is similar.

11.6   Exercise. A Prove that if $g(x)=\cos (rx)$, then $g^\prime (x)=-r\sin (rx).$

11.7   Theorem (Derivative of the logarithm.) The logarithm function is differentiable on $\mbox{${\mbox{{\bf R}}}^{+}$}$, and

$$\ln^\prime (x)={1\over x} \mbox{ for all } x\in\mbox{${\mbox{{\bf R}}}^{+}$}.$$

Proof: Let $x \in \mbox{${\mbox{{\bf R}}}^{+}$}$, and let $s\in\mbox{${\mbox{{\bf R}}}^{+}$}\setminus\{x\}$. Then

$${{\ln (s)-\ln (x)}\over {s-x}}={1\over {s-x}}\int_x^s {1\over t}dt={1\over {s-x}}A_x^s \Big[ {1\over t}\Big].$$

Case 1: If $s>x$ then $${A_x^s \Big[ {1\over t}\Big]}$$ represents the area of the shaded region $S$ in the figure.

We have

$$B(x,s\colon 0,{1\over s})\subset S\subset B(x,s\colon 0,{1\over x})$$

so by monotonicity of area

$${{s-x}\over s}\leq A_x^s\Big[ {1\over t}\Big]\leq {{s-x}\over x}.$$

Thus

$${1\over s}\leq {1\over {s-x}}\int_x^s {1\over t}dt\leq {1\over x}.$$ (11.8)

Case 2. If $s<x$ we can reverse the roles of $s$ and $x$ in equation (11.8) to get

$${1\over x}\leq {1\over {x-s}}\int_s^x {1\over t}dt\leq {1\over s}$$

or

$${1\over x}\leq {1\over {s-x}}\int_x^s {1\over t}dt\leq {1\over s}.$$

In both cases it follows that

$$0 \leq \Big\vert {1\over {s-x}}\int_x^s {1\over t}dt-{1\over x}\Big\vert\leq\Big\vert{1\over s}-{1\over x}\Big\vert.$$

Let $\{x_n\}$ be a generic sequence in $\mbox{${\mbox{{\bf R}}}^{+}$}\setminus\{x\}$ such that $\{x_n\}\to x$. Then $${\Big\{ {1\over {x_n}}-{1\over x}\Big\} \to 0}$$ , so by the squeezing rule

$$\left\{ {1\over {x_n-x}}\int_x^{x_n}{1\over t}dt-{1\over x} \right\} \to 0,$$

i.e.

$$\left\{ {{\ln (x_n)-\ln (x)}\over {x_n-x}}-{1\over x}\right\} \to 0.$$

Hence

$$\Big\{ {{\ln (x_n)-\ln (x)}\over {x_n-x}}\Big\}\to {1\over x}.$$

We have proved that $\ln^\prime (x)=\displaystyle {{1\over x}}$. $\diamondsuit$

11.9   Assumption (Localization rule for derivatives.) Let $f,g$ be two real valued functions. Suppose there is some $\epsilon\in\mbox{${\mbox{{\bf R}}}^{+}$}$ and $a\in\mbox{{\bf R}}$ such that

$$(a-\epsilon ,a+\epsilon )\subset{\rm domain}(f)\cap{\rm domain}(g)$$

and such that

$$f(x)=g(x) \mbox{ for all } x\in (a-\epsilon ,a+\epsilon).$$

If $f$ is differentiable at $a$, then $g$ is differentiable at $a$ and $g^\prime (a)=f^\prime (a)$.

This is another assumption that is really a theorem, i.e. it can be proved. Intuitively this assumption is very plausible. It says that if two functions agree on an entire interval centered at $a$, then their graphs have the same tangents at $a$.

11.10   Theorem (Derivative of absolute value.) Let $f(x)=\vert x\vert$ for all $x \in \mbox{{\bf R}}$. Then $${f^\prime (x)={x\over {\vert x\vert}}}$$ for all $x\in\mbox{{\bf R}}\setminus\{0\}$ and $f^\prime (0)$ is not defined.

Proof: Since

$$f(x)=\cases{ x &if $x>0$,\cr -x &if $x<0$,\cr}$$

it follows from the localization theorem that

$$f^\prime (x)=\cases{ 1={x\over {\vert x\vert}} &if $x>0$,\cr -1={x\over {\vert x\vert}} &if $x<0$.\cr}$$

To see that $f$ is not differentable at $0$, we want to show that

$$\lim_{t\to 0}{{f(t)-f(0)}\over {t-0}}=\lim_{t\to 0}{{\vert t\vert}\over t}$$

does not exist. Let $${x_n = {{(-1)^n}\over n}}$$. Then $\{x_n\} \to 0$, but $${ {{\vert x_n\vert}\over {x_n}}={{ \Big\vert {{(-1)^n}\over n}\Big\vert}\over {{(-1)^n}\over n}}=(-1)^n}$$ and we know that $\lim\{(-1)^n\}$ does not exist. Hence $${\lim {{f(t)-f(0)}\over {t-0}}}$$ does not exist, i.e., $f$ is not differentiable at $0$.

11.11   Definition ( $${ {d\over {dx}}}$$ notation for derivatives.) An alternate notation for representing derivatives is:

$${d\over {dx}}f(x)=f^\prime (x)$$

or

$${{df}\over {dx}}=f^\prime (x).$$

This notation is used in the following way

$$\begin{eqnarray*} {d\over {dx}}\Big(\sin (6x)\Big) &=& 6\cos (6x),\\ {d\over {dt}}\Big(\cos ({t\over 3})\Big) &=& -{1\over 3}\sin ( {t\over 3}). \end{eqnarray*}$$

Or:

Let $f=x^{1/2}$. Then $${ {{df}\over {dx}}={1\over 2}x^{-1/2}}$$.

Let $g(x)=\displaystyle {{1\over x}}$. Then $${ {{dg}\over {dx}}={d\over {dx}}\Big( g(x)\Big)={d\over {dx}}\Big( {1\over x}\Big)=-{1\over {x^2}}}$$.

The $${ {d\over {dx}}}$$ notation is due to Leibnitz, and is older than our concept of function.

Leibnitz wrote the differentiation formulas as `` $dx^a=ax^{a-1}dx$," or if $y=x^a$, then `` $dy=ax^{a-1}dx.$" The notation $f^\prime (x)$ for derivatives is due to Joseph Louis Lagrange (1736-1813). Lagrange called $f^\prime (x)$ the derived function of $f(x)$ and it is from this that we get our word derivative. Leibnitz called derivatives, differentials and Newton called them fluxions.

Many of the early users of the calculus thought of the derivative as the quotient of two numbers

$${df \over dx} = {\mbox{difference in $f$} \over \mbox{difference in $x$} } = {f(x) - f(t) \over x - t}$$

when $dx= x-t$ was ``infinitely small''. Today ``infinitely small'' real numbers are out of fashion, but some attempts are being made to bring them back. Cf Surreal Numbers : How two ex-students turned on to pure mathematics and found total happiness : a mathematical novelette, by D. E. Knuth.[30]. or The Hyperreal Line by H. Jerome Keisler[28, pp 207-237].