[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.