11.2 Some General Differentiation Theorems.

11.12   Theorem (Sum rule for derivatives.) Let $f,g$ be real valued functions with domain $(f)\subset\mbox{{\bf R}}$ and domain $(g)\subset\mbox{{\bf R}}$, and let $c \in \mbox{{\bf R}}$. Suppose $f$ and $g$ are differentiable at $a$. Then $f+g$, $f-g$ and $cf$ are differentiable at $a$, and

$$\begin{eqnarray*} (f+g)^\prime (a) &=& f^\prime (a)+g^\prime (a)\\ (f-g)^\prime (a) &=& f^\prime (a)-g^\prime (a) \\ (cf)'(a) &=& c\cdot f'(a). \end{eqnarray*}$$

Proof: We will prove only the first statement. The proofs of the other statements are similar. For all $x \in \mbox{{\rm dom}}(f)$ we have

$$\begin{eqnarray*} {(f+g)(x) - (f+g)(a) \over x-a} &=& {f(x) + g(x) - ( f(a) + g(... ...r x-a} \\ &=& {f(x) - f(a) \over x-a} + {g(x) - g(a) \over x-a} \end{eqnarray*}$$

By the sum rule for limits of functions, it follows that

$$\lim_{x\to a}\left( {(f+g)(x) - (f+g)(a) \over x-a} \right) =... ...\right) + \lim_{x\to a}\left( {g(x) - g(a) \over x-a} \right),$$

i.e.

$$(f+g)'(a) = f'(a) + g'(a).\mbox{ $\diamondsuit$}$$

11.13   Examples. If

$$f(x)=27x^3+{1\over {3x}}+\sqrt{8x},$$

then

$$f(x)=27x^3+{1\over 3}x^{-1}+\sqrt 8 \cdot x^{1/2},$$

so

$$\begin{eqnarray*} f^\prime (x) &=& 27\cdot (3x^2)+{1\over 3}(-1\cdot x^{-2})+\sq... ... 2}x^{-1/2}) \\ &=& 81x^2 -{1\over {3x^2}}+ \sqrt{ {2\over x}}. \end{eqnarray*}$$

If $g(x)=(3x^2+7)^2$, then $g(x)=9x^4+42x^2+49$, so

$$g^\prime (x)=9\cdot 4x^3+42\cdot 2x=36x^3+84x.$$

If $h(x)=\sin (4x)+\sin^2(4x)$, then $${h(x)=\sin (4x)+{1\over 2}\Big(1-\cos (8x)\Big)}$$, so

$$\begin{eqnarray*} h^\prime (x) &=& 4\cos (4x)+{1\over 2}(-1)\Big( -8\cdot\sin (8x)\Big)\\ &=& 4\cos (4x) +4\sin (8x). \end{eqnarray*}$$

$${d\over {ds}}\Big(8\sin (4s)+s^2+4\Big)=32\cos (4s)+2s.$$

11.14   Exercise. Calculate the derivatives of the following functions:

a)

$f(x)=(x^2+4x)^2$

b)

$${ g(x)=\sqrt{3x^3}+{4\over{x^4}}}$$

c)

$h(t)=\ln (t)+\ln (t^2)+\ln (t^3)$

d)

$k(x)=\ln (10\cdot x^{5/2})$

e)

$l(x)=3\cos (x)+\cos (3x)$

f)

$m(x)=\cos (x)\cos (3x)$

g)

$n(x)=\Big(\sin^2(x)+\cos^2(x)\Big)^4$

11.15   Exercise. A Calculate

a)

$${ {d\over {dt}}\Big( 1+t+{{t^2}\over {2!}}+{{t^3}\over {3!}}+{{t^4}\over {4!}}\Big)}$$

b)

$${ {d\over {dt}}\Big( h_0+v_0(t-t_0)-{1\over 2}g(t-t_0)^2\Big)}$$. Here $h_0,v_0,t_0$ and $g$ are all constants.

c)

$${ {d\over {dt}}(\vert-100t\vert)}$$

11.16   Theorem (The product rule for derivatives.) Let $f$ and $g$ be real valued functions with $\mbox{{\rm dom}}(f)\subset\mbox{{\bf R}}$ and $\mbox{{\rm dom}}(g)\subset\mbox{{\bf R}}$. Suppose $f$ and $g$ are both differentiable at $a$. Then $fg$ is differentiable at $a$ and

$$(fg)^\prime (a)=f(a)\cdot g^\prime (a)+f^\prime (a)\cdot g(a).$$

In particular, if $f= c$ is a constant function, we have

$$(cf)'(a) = c\cdot f'(a).$$

Proof: Let $x$ be a generic point of $\mbox{{\rm dom}}(f)\cap\mbox{{\rm dom}} (g)\setminus\{a\}$. Then

$$\begin{eqnarray*} {{(fg)(x)-(fg)(a)}\over {x-a}} &=& {{f(x)\Big( g(x)-g(a)\Big)+... ...)+\Bigg( {{f(x)-f(a)}\over {x-a}}\Bigg) g(a). \mbox{{\nonumber}} \end{eqnarray*}$$

We know that $${\lim_{x\to a}\Big( {{g(x)-g(a)}\over {x-a}}\Big)=g^\prime (a)}$$ and $${\lim_{x\to a}\Big( {{f(x)-f(a)}\over {x-a}}\Big) =f^\prime (a)}$$. If we also knew that $${\lim_{x\to a}f(x)=f(a)}$$, then by basic properties of limits we could say that

$$(fg)^\prime (a)=\lim_{x\to a}{{(fg)(x)-(fg)(a)}\over {x-a}}=f(a)g^\prime (a)+f^\prime (a)g(a)$$

which is what we claimed.

This missing result will be needed in some other theorems, so I've isolated it in the following lemma.

11.17   Lemma (Differentiable functions are continuous.) Let $f$ be a real valued function such that $\mbox{{\rm dom}}(f)\subset\mbox{${\mbox{{\bf R}}}^{+}$}$. Suppose $f$ is differentiable at a point $a\in\mbox{{\rm dom}}(f)$. Then $${\lim_{x\to a}f(x)=f(a)}$$. (We will define `` continuous" later. Note that neither the statement nor the proof of this lemma use the word `` continuous" in spite of the name of the lemma.)

Proof:

$$\lim_{x\to a}f(x)=\lim_{x\to a}\Bigg( {{f(x)-f(a)}\over {x-a}}\cdot (x-a)+f(a)\Bigg).$$

Hence by the product and sum rules for limits,

$$\lim_{x\to a}f(x)=f^\prime (a)\cdot (a-a)+f(a)=f(a).\mbox{ $\diamondsuit$}$$

11.18   Example (Leibniz's proof of the product rule.) Leibniz stated the product rule as

$$dxy = xdy + ydx\cite[page 143]{leibchild.b}\footnotemark$$

His proof is as follows:

$dxy$ is the difference between two successive $xy$'s; let one of these be $xy$ and the other $x+dx$ into $y+dy$; then we have

$$dxy = \overline{\overline{x+dx} \cdot \overline{y+dy}} - xy = xdy + ydx + dxdy;$$

the omission of the quantity $dxdy$ which is infinitely small in comparison with the rest, for it is supposed that $dx$ and $dy$ are infinitely small (because the lines are understood to be continuously increasing or decreasing by very small increments throughout the series of terms), will leave $xdy + ydx$.[34, page 143]

Notice that for Leibniz, the important thing is not the derivative, $${dxy \over dt}$$, but the infinitely small differential, $dxy$.

11.19   Theorem (Derivative of a reciprocal.) Let $f$ be a real valued function such that $\mbox{{\rm dom}}(f)\subset\mbox{{\bf R}}$. Suppose $f$ is differentiable at some point $a$, and $f(a)\neq 0$. Then $${{1\over f}}$$ is differentiable at $a$, and

$$\Big( {1\over f}\Big)^\prime (a)={{-f^\prime (a)}\over {\Big( f(a)\Big)^2}}.$$

Proof: For all $${ x\in\mbox{{\rm dom}}\Big( {1\over f}\Big)\setminus\{a\} }$$

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

It follows from the standard limit rules that

$$\lim_{x\to a} {{ {1\over {f(x)}}-{1\over {f(a)}}}\over {x-a}}=-f^\prime (a)\cdot {1\over {\Big( f(a)\Big)^2}}.$$

11.20   Theorem (Quotient rule for derivatives.) Let $f,g$ be real valued
functions with $\mbox{{\rm dom}}(f)\subset\mbox{{\bf R}}$ and $\mbox{{\rm dom}}(g)\subset\mbox{{\bf R}}$. Suppose $f$ and $g$ are both differentiable at $a$, and that $g(a)\neq 0$. Then $${ {f\over g}}$$ is differentiable at $a$, and

$$\Big( {f\over g}\Big)^\prime (a)={{g(a)f^\prime (a)-f(a)g^\prime (a)}\over {\Big(g(a)\Big)^2}}.$$

11.21   Exercise. A Prove the quotient rule.

11.22   Examples. Let

$$f(x)={{\sin (x)}\over x} \mbox{ for } x\in\mbox{{\bf R}}\setminus\{0\}.$$

Then by the quotient rule

$$f^\prime (x)={{x(\cos (x))-\sin (x)}\over {x^2}}.$$

Let $h(x)=x^2\cdot\vert x\vert$. Then by the product rule

$$h^\prime (x)=x^2 \Big( {x\over {\vert x\vert}}\Big) +2x\vert x\vert=x\vert x\vert+2x\vert x\vert=3x\vert x\vert$$

(since $${ {{x^2}\over {\vert x\vert}}={{\vert x\vert^2}\over {\vert x\vert}}=\vert x\vert}$$).

The calculation is not valid at $x=0$ (since $\vert x\vert$ is not differentiable at $0$, and we divided by $\vert x\vert$ in the calculation. However $h$ is differentiable at $0$ since $${ \lim_{t\to 0}{{h(t)-h(0)}\over {t-0}}=\lim_{t\to 0} {{t^2\vert t\vert}\over t}=\lim_{t\to 0} t\vert t\vert=0}$$, i.e., $h^\prime (0)=0=3\cdot 0\cdot \vert\vert$. Hence the formula

$${d\over {dx}}(x^2\vert x\vert)=3x\vert x\vert$$

is valid for all $x \in \mbox{{\bf R}}$.

Let $g(x)=\ln (x)\cdot\sin (10x)\cdot\sqrt x$. Consider $g$ to be a product $g=hk$ where $h(x)=\ln (x)\cdot\sin (10x)$ and $k(x)=\sqrt x$. Then we can apply the product rule twice to get

$$\begin{eqnarray*} g^\prime (x) &=& \Big( \ln (x)\cdot\sin (10x)\Big)\cdot {1\ove... ...x)\cdot \Big(10\cos (10x)\Big)+{1\over x}\sin (10x)\Bigg)\sqrt x.\end{eqnarray*}$$

11.23   Exercise (Derivatives of tangent, cotangent, secant, cosecant.) We define functions tan, cot, sec, and csc by

$$\begin{array}{ll} \tan (x)=\displaystyle {{\sin (x)}\over {\c... ...x)}}, & \csc (x)=\displaystyle {1\over {\sin (x)}}. \end{array}$$

The domains of these functions are determined by the definition of the domain of a quotient, e.g. $\mbox{{\rm dom}}(\sec)=\{x\in\mbox{{\bf R}}\colon\cos x\neq 0\}$. Prove that

$$\begin{array}{ll} {d\over {dx}}\tan (x)=\sec^2(x), & {d\over ... ...sec (x), & {d\over {dx}}\csc (x)=-\cot (x)\csc (x). \end{array}$$

(You should memorize these formulas. Although they are easy to derive, later we will want to use them backwards; i.e., we will want to find a function whose derivative is $\sec^2 (x)$. It is not easy to derive the formulas backwards.)

11.24   Exercise. A Calculate the derivatives of the following functions. Simplify your answers if you can.

a)

$f(x)=x\cdot \ln (x)-x$.

b)

$g(x)=\displaystyle { {{ax+b}\over {cx +d}}}$ (here $a, b, c, d$ are constants).

c)

$k(x)=(x^2+3x+10)(x^2+3x+12)$.

d)

$${ m(x)= {{\cos (6x)}\over {\cos (7x)}}}$$.

11.25   Exercise. Let $f$, $g$, $h$, and $k$ be differentiable functions defined on $\mbox{{\bf R}}$.

a) Express $(fgh)'$ in terms of $f$, $f'$, $g$, $g'$, $h$ and $h'$.

b) On the basis of your answer for part a), try to guess a formula for $(fghk)'$. Then calculate $(fghk)'$, and see whether your guess was right.