[Calculus] IV. Differentiation

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Differentiation Formulas

Theorem 4.1.1.

$\displaystyle (1) \hspace{2.5em} {\operatorname{d}\over\operatorname{d}\!x}(c) = 0 \\[14pt] \displaystyle (2) \hspace{2.5em} {\operatorname{d}\over\operatorname{d}\!x}(x^{n}) = nx^{n - 1} \qquad \left( n \in \mathbb{R} \right) \\[14pt] \displaystyle (3) \hspace{2.5em} {\operatorname{d}\over\operatorname{d}\!x}\!\left[ cf(x) \right] = c{\operatorname{d}\over\operatorname{d}\!x}f(x)$

$\displaystyle (4) \hspace{3em}$If $f$ and $g$ are both differentiable, then

$\displaystyle \hspace{3.8em} {\operatorname{d}\over\operatorname{d}\!x}\!\left[ f(x) + g(x) \right] = {\operatorname{d}\over\operatorname{d}\!x}f(x) + {\operatorname{d}\over\operatorname{d}\!x}g(x)$

$\displaystyle \hspace{3.8em} {\operatorname{d}\over\operatorname{d}\!x}\!\left[ f(x) - g(x) \right] = {\operatorname{d}\over\operatorname{d}\!x}f(x) - {\operatorname{d}\over\operatorname{d}\!x}g(x)$

$\displaystyle (5) \hspace{3em}$If $f$ and $g$ are both differentiable, then

$\displaystyle \hspace{3.8em} {\operatorname{d}\over\operatorname{d}\!x}\!\left[ f(x) g(x) \right] = f(x){\operatorname{d}\over\operatorname{d}\!x}g(x) + g(x){\operatorname{d}\over\operatorname{d}\!x}f(x)$

$\displaystyle \hspace{3.8em} {\operatorname{d}\over\operatorname{d}\!x}\!\left[ \frac{f(x)}{g(x)} \right] = \frac{ \displaystyle g(x){\operatorname{d}\over\operatorname{d}\!x}f(x) - f(x){\operatorname{d}\over\operatorname{d}\!x}g(x)}{ \displaystyle \left[ g(x) \right]^{2}}$

Proof.  (3) Let $g(x) = cf(x)$. Then

$\displaystyle \begin{align} {\operatorname{d}\over\operatorname{d}\!x}g(x) &= \lim_{h \to 0}\frac{g(x + h) - g(x)}{h} \\[7pt] &= \lim_{h \to 0}\frac{cf(x + h) - cf(x)}{h} \\[7pt] &= \lim_{h \to 0}c\left[ \frac{f(x + h) - f(x)}{h} \right] \\[7pt] &= c\lim_{h \to 0}\frac{f(x + h) - f(x)}{h} \\[7pt] &= c{\operatorname{d}\over\operatorname{d}\!x}f(x) \end{align}$

$\Box$

(4)(1) Let $F(x) = f(x) + g(x)$. Then

$\displaystyle \begin{align} {\operatorname{d}\over\operatorname{d}\!x}F(x) &= \lim_{h \to 0}\frac{F(x + h) - F(x)}{h} \\[7pt] &= \lim_{h \to 0}\frac{\left[ f(x + h) + g(x + h) \right] - \left[ f(x) + g(x) \right]}{h} \\[7pt] &= \lim_{h \to 0}\left[ \frac{f(x + h) - f(x)}{h} + \frac{g(x + h) - g(x)}{h} \right] \\[7pt] &= {\operatorname{d}\over\operatorname{d}\!x}f(x) + {\operatorname{d}\over\operatorname{d}\!x}g(x) \end{align}$

$\Box$

(5)(1) Let $F(x) = f(x)g(x)$. Then

$\displaystyle \begin{align} {\operatorname{d}\over\operatorname{d}\!x}F(x) &= \lim_{h \to 0}\frac{F(x + h) - F(x)}{h} \\[7pt] &= \lim_{h \to 0}\frac{f(x + h) g(x + h) - f(x) g(x)}{h} \end{align}$

In order to evaluate this limit, we would like to separate the functions $f$ and $g$ as in the proof of (4)(1). We can achieve this separation by subtracting and adding the term $f(x + h)g(x)$ in the numerator:

$\begin{align} \displaystyle &{\operatorname{d}\over\operatorname{d}\!x}F(x) \\[14pt] \displaystyle &= \lim_{h \to 0}\frac{f(x + h) g(x + h) - f(x + h)g(x) + f(x + h)g(x) - f(x) g(x)}{h} \\[14pt] \displaystyle &= \lim_{h \to 0}\!\left[ f(x + h)\frac{g(x + h) - g(x)}{h} + g(x)\frac{f(x + h) - f(x)}{h} \right] \\[14pt] \displaystyle &= \lim_{h \to 0}f(x + h) \cdot \lim_{h \to 0}\frac{g(x + h) - g(x)}{h} + \lim_{h \to 0}g(x) \cdot \lim_{h \to 0}\frac{f(x + h) - f(x)}{h} \\[14pt] \displaystyle &= f(x){\operatorname{d}\over\operatorname{d}\!x}g(x) + g(x){\operatorname{d}\over\operatorname{d}\!x}f(x) \end{align}$

$\Box$

(5)(2) Let $\displaystyle F(x) = \frac{f(x)}{g(x)}$. Then

$\displaystyle \begin{align} {\operatorname{d}\over\operatorname{d}\!x}F(x) &= \lim_{h \to 0}\frac{F(x + h) - F(x)}{h} \\[7pt] &= \lim_{h \to 0}\frac{ \displaystyle \frac{f(x + h)}{g(x + h)} - \frac{f(x)}{g(x)}}{h} \\[7pt] &= \lim_{h \to 0}\frac{f(x + h)g(x) - f(x)g(x + h)}{hg(x + h)g(x)} \end{align}$

We can separate $f$ and $g$ in this expression by subtracting and adding the term $f(x)g(x)$ in the numerator:

$\begin{align} \displaystyle &{\operatorname{d}\over\operatorname{d}\!x}F(x) \\[14pt] \displaystyle &= \lim_{h \to 0}\frac{f(x + h)g(x) - f(x)g(x) + f(x)g(x) - f(x)g(x + h)}{hg(x + h)g(x)} \\[14pt] \displaystyle &= \lim_{h \to 0}\frac{\displaystyle g(x)\frac{f(x + h) - f(x)}{h} - f(x)\frac{g(x + h) - g(x)}{h}}{g(x + h)g(x)} \\[14pt] \displaystyle &= \frac{\displaystyle \lim_{h \to 0}g(x) \cdot \lim_{h \to 0}\frac{f(x + h) - f(x)}{h} - \displaystyle \lim_{h \to 0}f(x) \cdot \lim_{h \to 0}\frac{g(x + h) - g(x)}{h}}{\displaystyle \lim_{h \to 0}g(x + h) \cdot \lim_{h \to 0}g(x)} \\[14pt] \displaystyle &= \frac{\displaystyle g(x){\operatorname{d}\over\operatorname{d}\!x}f(x) - f(x){\operatorname{d}\over\operatorname{d}\!x}g(x)}{\left[ g(x) \right]^{2}} \end{align}$

$\Box$

Example 4.1.2.  Find $F'(x)$ if $F(x) = \left(6x^{3}\right)\!\left(7x^{4}\right)$.

Solution.  By Theorem 4.1.1 - (5)(1), we have

$\displaystyle \begin{align} F'(x) &= \left(6x^{3}\right)\!{\operatorname{d}\over\operatorname{d}\!x}\!\left(7x^{4}\right) + \left(7x^{4}\right)\!{\operatorname{d}\over\operatorname{d}\!x}\!\left(6x^{3}\right) \\[7pt] &= \left(6x^{3}\right)\!\left(28x^{3}\right) + \left(7x^{4}\right)\!\left(18x^{2}\right) \\[7pt] &= 168x^{6} + 126x^{6} = 294x^{6} \end{align}$

$\Box$

Lemma 4.1.3.

$\displaystyle \lim_{\theta \to 0}\frac{\sin \theta}{\theta} = 1$

Proof.  We now use a geometric argument to prove the limit above. Assume first that $\theta$ lies between $0$ and $\pi/2$.

Figure 1

Figure 2

Figure 1 shows a sector of a circle with center $O$, central angle $\theta$, and radius $1$. $BC$ is drawn perpendicular to $OA$. By the definition of radian measure, we have arc $AB = \theta$. Also $\left\vert BC \right\vert = \left\vert OB \right\vert \sin \theta$ $= \sin \theta$. From the diagram we see that

$\displaystyle \left\vert BC \right\vert < \left\vert AB \right\vert < \text{arc} \ AB$

Therefore

$\displaystyle \sin \theta < \theta \qquad \text{so} \qquad \frac{\sin \theta}{\theta} < 1$

Let the tangent lines at $A$ and $B$ intersect at $E$. You can see from Figure 2 that the circumference of a circle is smaller than the length of a circumscribed polygon, and so arc $AB < \left\vert AE \right\vert + \left\vert EB \right\vert$. Thus

$\displaystyle \begin{align} \theta = \text{arc} \ AB &< \left\vert AE \right\vert + \left\vert EB \right\vert \\[7pt] &< \left\vert AE \right\vert + \left\vert ED \right\vert \\[7pt] &= \left\vert AD \right\vert = \left\vert OA \right\vert \tan \theta \\[7pt] &= \tan \theta \end{align}$

Therefore we have

$\displaystyle \theta < \frac{\sin \theta}{\cos \theta}$

so

$\displaystyle \cos \theta < \frac{\sin \theta}{\theta} < 1$

We know $\displaystyle \lim_{\theta \to 0}1 = 1$ and $\displaystyle \lim_{\theta \to 0}\cos \theta = 1$, so by Theorem 1.2.3, we have

$\displaystyle \lim_{\theta \to 0^{+}}\! \! \frac{\sin \theta}{\theta} = 1$

But the function $(\sin \theta)/\theta$ is an even function, so its right and left limits must be equal. Hence, we have

$\displaystyle \lim_{\theta \to 0}\frac{\sin \theta}{\theta} = 1$

$\Box$

Theorem 4.1.4.

$\displaystyle (1) \hspace{2.5em} {\operatorname{d}\over\operatorname{d}\!x}(\sin x) = \cos x \\[14pt] \displaystyle (2) \hspace{2.5em} {\operatorname{d}\over\operatorname{d}\!x}(\cos x) = -\sin x \\[14pt] \displaystyle (3) \hspace{2.5em} {\operatorname{d}\over\operatorname{d}\!x}(\tan x) = \sec^2 \! x \\[14pt] \displaystyle (4) \hspace{2.5em} {\operatorname{d}\over\operatorname{d}\!x}(\csc x) = -\csc x \cot x \\[14pt] \displaystyle (5) \hspace{2.5em} {\operatorname{d}\over\operatorname{d}\!x}(\sec x) = \sec x \tan x \\[14pt] \displaystyle (6) \hspace{2.5em} {\operatorname{d}\over\operatorname{d}\!x}(\cot x) = -\cot^2 \! x \\[14pt]$

Proof.  (1) Let $f(x) = \sin x$. Then

$\displaystyle \begin{align} {\operatorname{d}\over\operatorname{d}\!x}f(x) &= \lim_{h \to 0}\frac{f(x + h) - f(x)}{h} \\[7pt] &= \lim_{h \to 0}\frac{\sin (x + h) - \sin x}{h} \\[7pt] &= \lim_{h \to 0}\frac{\sin x \cos h + \cos x \sin h - \sin x}{h} \\[7pt] &= \lim_{h \to 0}\left[ \frac{\sin x(\cos h - 1)}{h} + \frac{\cos x \sin h}{h} \right] \\[7pt] &= \lim_{h \to 0}\left[ \sin x \left(\frac{\cos h - 1}{h} \right) + \cos x \left(\frac{\sin h}{h} \right) \right] \\[7pt] &= \lim_{h \to 0}\sin x \cdot \lim_{h \to 0}\frac{\cos h - 1}{h} + \lim_{h \to 0}\cos x \cdot \lim_{h \to 0}\frac{\sin h}{h} \end{align}$

By Lemma 4.1.3, $\displaystyle \lim_{h \to 0}(\sin h)/h = 1$. We can deduce the value of $\displaystyle \lim_{h \to 0}$$(\cos h - 1)/h$ as follows:

$\displaystyle \begin{align} \lim_{h \to 0}\frac{\cos h - 1}{h} &= \lim_{h \to 0}\left( \frac{\cos h - 1}{h} \cdot \frac{\cos h + 1}{\cos h + 1} \right) = \lim_{h \to 0}\frac{\cos^2 \! h - 1}{h(\cos h + 1)} \\[7pt] &= \lim_{h \to 0}\frac{-\sin^2 \! h}{h(\cos h + 1)} = -\lim_{h \to 0}\left( \frac{\sin h}{h} \cdot \frac{\sin h}{\cos h + 1} \right) \\[7pt] &= -\lim_{h \to 0}\frac{\sin h}{h} \cdot \lim_{h \to 0}\frac{\sin h}{\cos h + 1} \\[7pt] &= -1 \cdot \left( \frac{0}{1 + 1} \right) = 0 \end{align}$

Therefore,

$\displaystyle \begin{align} {\operatorname{d}\over\operatorname{d}\!x}f(x) &= \lim_{h \to 0}\sin x \cdot \lim_{h \to 0}\frac{\cos h - 1}{h} + \lim_{h \to 0}\cos x \cdot \lim_{h \to 0}\frac{\sin h}{h} \\[7pt] &= (\sin x) \cdot 0 + (\cos x) \cdot 1 = \cos x \end{align}$

$\Box$

(3)

$\displaystyle \begin{align} {\operatorname{d}\over\operatorname{d}\!x}(\tan x) &= {\operatorname{d}\over\operatorname{d}\!x}\!\left( \frac{\sin x}{\cos x} \right) \\[7pt] &= \frac{\displaystyle \cos x {\operatorname{d}\over\operatorname{d}\!x}(\sin x) - \sin x {\operatorname{d}\over\operatorname{d}\!x}(\cos x)}{\cos^2 \! x} \\[7pt] &= \frac{\cos x \cdot \cos x - \sin x (-\sin x)}{\cos^2 \! x} \\[7pt] &= \frac{\cos^2 \! x + \sin^2 \! x}{\cos^2 \! x} \\[7pt] &= \frac{1}{\cos^2 \! x} = \sec^2 \! x \end{align}$

$\Box$

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The Chain Rule

Theorem 4.2.1. (The Chain Rule)  If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$, then the composite function $F = f \circ g$ defined by $F(x) = f(g(x))$ is differentiable at $x$ and $F'$ is given by the product

$\displaystyle F'(x) = f'(g(x)) \cdot g'(x)$

In Leibniz notation, if $y = f(u)$ and $u = g(x)$ are both differentiable functions, then

$\displaystyle {\operatorname{d}\!y\over\operatorname{d}\!x} = {\operatorname{d}\!y\over\operatorname{d}\!u} {\operatorname{d}\!u\over\operatorname{d}\!x}$

Proof.  Recall that if $y = f(x)$ and $x$ changes from $a$ to $a + \Delta x$, we define the increment of $y$ as

$\displaystyle \Delta y = f(a + \Delta x) - f(a)$

According to the definition of a derivative, we have

$\displaystyle \lim_{\Delta x \to 0}\! \frac{\Delta y}{\Delta x} = f'(a)$

So if we denote by $\varepsilon$ the difference between the difference quotient and the derivative, we obtain

$\displaystyle \lim_{\Delta x \to 0}\! \varepsilon = \lim_{\Delta x \to 0}\!\! \left( \frac{\Delta y}{\Delta x} - f'(a) \right) = f'(a) - f'(a) = 0$

But

$\displaystyle \varepsilon = \frac{\Delta y}{\Delta x} - f'(a) \qquad \Rightarrow \qquad \Delta y = f'(a)\Delta x + \varepsilon \Delta x$

If we define $\varepsilon$ to be $0$ when $\Delta x = 0$, then $\varepsilon$ becomes a continuous function of $\Delta x$. Thus, for a differentiable function $f$, we can write

$\displaystyle \Delta y = f'(a)\Delta x + \varepsilon \Delta x \qquad \text{where} \qquad \varepsilon \to 0 \quad \text{as} \quad \Delta x \to 0 \qquad \cdots \qquad \text{(i)}$

and $\varepsilon$ is a continuous function of $\Delta x$. This property of differentiable functions is what enables us to prove the Chain Rule.

Suppose $u = g(x)$ is differentiable at $a$ and $y = f(u)$ is differentiable at $b = g(a)$. If $\Delta x$ is an increment in $x$ and $\Delta u$ and $\Delta y$ are the corresponding increments in $u$ and $y$, then we can use Equation $(\text{i})$ to write

$\displaystyle \Delta u = g'(a)\Delta x + \varepsilon_{1}\Delta x = \left[ g'(a) + \varepsilon_{1} \right]\Delta x \qquad \cdots \qquad \text{(ii)}$

where $\varepsilon_{1} \to 0$ as $\Delta x \to 0$. Similarly

$\displaystyle \Delta y = f'(b)\Delta u + \varepsilon_{2}\Delta u = \left[ f'(b) + \varepsilon_{2} \right]\Delta u \qquad \cdots \qquad \text{(iii)}$

where $\varepsilon_{2} \to 0$ as $\Delta u \to 0$. If we now substitute the expression for $\Delta u$ from Equation $(\text{ii})$ into Equation $(\text{iii})$, we get

$\displaystyle \Delta y = \left[ f'(b) + \varepsilon_{2} \right] \left[ g'(a) + \varepsilon_{1} \right]\Delta x$

so

$\displaystyle \frac{\Delta y}{\Delta x} = \left[ f'(b) + \varepsilon_{2} \right] \left[ g'(a) + \varepsilon_{1} \right]$

As $\Delta x \to 0$, Equation $(\text{ii})$ shows that $\Delta u \to 0$. So both $\varepsilon_{1} \to 0$ and $\varepsilon_{2} \to 0$ as $\Delta x \to 0$. Therefore

$\displaystyle \begin{align} {\operatorname{d}\!y\over\operatorname{d}\!x} &= \lim_{\Delta x \to 0}\!\! \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0}\! \left[ f'(b) + \varepsilon_{2} \right] \left[ g'(a) + \varepsilon_{1} \right] \\[7pt] &= f'(b)g'(a) = f'(g(a))g'(a) \end{align}$

This proves the Chain Rule.

$\Box$

Example 4.2.2.  Differentiate $y = \sin \left(x^{2}\right)$.

Solution.  Let $u = x^{2}$. Then by Theorem 4.2.1,

$\displaystyle \begin{align} {\operatorname{d}\!y\over\operatorname{d}\!x} &= {\operatorname{d}\!y\over\operatorname{d}\!u}{\operatorname{d}\!u\over\operatorname{d}\!x} = {\operatorname{d}\over\operatorname{d}\!u}\!\sin u \cdot {\operatorname{d}\over\operatorname{d}\!x}x^{2} \\[7pt] &= \cos u \cdot 2x = 2x\cos \left(x^{2}\right) \end{align}$

$\Box$

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The body text and the figures were excerpted from James Stewart, Calculus 8E.