Lesson 31: Product and Quotient Rules II

Objectives

Use basic rules plus the product and quotient rules to find derivatives by hand.
Using R, obtain and plot the derivative of a function.

1. Derivatives by Hand

Objective. Use basic rules plus the product and quotient rules to find derivatives by hand.

Order of work. Finding a derivative by hand combines the basic rules from sections 2 to 4 of the Lesson 29 Notes with the product and quotient rules from section 2 of the Lesson 30 Notes.

Rewrite roots and reciprocals of powers of $x$ as powers, as in $\sqrt{x} = x^{1/2}$ and $1/x^2 = x^{-2}$ .
Identify the main operation: a sum, a constant multiple, a product, or a quotient.
For a sum, differentiate term by term. For a constant multiple, keep the constant and differentiate the function it multiplies. Apply steps 4 and 5 to each term or function that is a product or a quotient.
For a product or a quotient, name $u$ and $v$ : the factors of a product, or the numerator and denominator of a quotient.
Find $u'$ and $v'$ . When $u$ or $v$ is itself a product or a quotient, find its derivative by the same steps. Substitute $u$ , $u'$ , $v$ , and $v'$ , each in parentheses, into the product rule or the quotient rule.
Simplify the result.

Example 1

For the function

y = \frac{x^4 - 3x^2 + 6}{x^2},

find $y'$ by simplifying first. Then find $y'$ by the quotient rule.

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To simplify first, divide each term of the numerator by $x^2$ :

y = \frac{x^4}{x^2} - \frac{3x^2}{x^2} + \frac{6}{x^2} = x^2 - 3 + 6x^{-2}.

Differentiate term by term with the power, constant multiple, and sum rules:

y' = 2x - 0 + 6(-2)x^{-3} = 2x - 12x^{-3} = 2x - \frac{12}{x^3}.

For the quotient rule, find the derivatives of the numerator $u$ and the denominator $v$ :

u = x^4 - 3x^2 + 6, \qquad u' = 4x^3 - 6x, \qquad v = x^2, \qquad v' = 2x.

Substitute $u$ , $u'$ , $v$ , and $v'$ into the quotient rule, $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$ , and expand the numerator:

y' = \frac{(4x^3 - 6x)x^2 - (x^4 - 3x^2 + 6)(2x)}{(x^2)^2} = \frac{4x^5 - 6x^3 - (2x^5 - 6x^3 + 12x)}{x^4} = \frac{2x^5 - 12x}{x^4} = 2x - \frac{12}{x^3}.

The two methods give the same derivative, and simplifying first takes fewer steps.

Example 2

Find the derivative:

y = 4e^x 3^x - 2\ln(x) + 5x^3.

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The main operation is a sum. Its first term is 4 times the product $e^x 3^x$ . Differentiate the product by substituting $u = e^x$ , $u' = e^x$ , $v = 3^x$ , and $v' = 3^x\ln(3)$ into the product rule, $(uv)' = u'v + uv'$ :

\frac{d}{dx}\big(e^x 3^x\big) = e^x 3^x + e^x 3^x\ln(3) = e^x 3^x\big(1 + \ln(3)\big).

Differentiate the whole sum term by term with the constant multiple, logarithm, and power rules:

y' = 4e^x 3^x\big(1 + \ln(3)\big) - 2 \cdot \frac{1}{x} + 5 \cdot 3x^2 = 4e^x 3^x\big(1 + \ln(3)\big) - \frac{2}{x} + 15x^2.

Example 3

Find the derivative:

f(x) = \frac{x^5 - \sqrt{x}}{\cos(x)}.

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The main operation is a quotient. Rewrite $\sqrt{x}$ as $x^{1/2}$ , and write the numerator $u$ and the denominator $v$ with their derivatives:

u = x^5 - x^{1/2}, \qquad u' = 5x^4 - \tfrac{1}{2}x^{-1/2}, \qquad v = \cos(x), \qquad v' = -\sin(x).

Substitute $u$ , $u'$ , $v$ , and $v'$ into the quotient rule. Since $v' = -\sin(x)$ , the term $-uv'$ becomes $+\bigl(x^5 - \sqrt{x}\bigr)\sin(x)$ in the numerator. In the second line, $x^{1/2}$ and $x^{-1/2}$ are written as roots again:

\begin{align*} f'(x) &= \frac{\left(5x^4 - \tfrac{1}{2}x^{-1/2}\right)\cos(x) - \left(x^5 - x^{1/2}\right)\big(-\sin(x)\big)}{\cos^2(x)}\\ &= \frac{\left(5x^4 - \dfrac{1}{2\sqrt{x}}\right)\cos(x) + \left(x^5 - \sqrt{x}\right)\sin(x)}{\cos^2(x)}. \end{align*}

Example 4

Find the derivative:

y = \frac{\cos(x)}{(x^3 + 1)\,e^x}.

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The main operation is a quotient with $u = \cos(x)$ , $u' = -\sin(x)$ , and $v = (x^3 + 1)e^x$ . The denominator is itself a product. Find $v'$ by the product rule, with factors $x^3 + 1$ and $e^x$ and their derivatives $3x^2$ and $e^x$ :

v' = 3x^2 e^x + (x^3 + 1)e^x = e^x(x^3 + 3x^2 + 1).

Substitute $u$ , $u'$ , $v$ , and $v'$ into the quotient rule:

y' = \frac{\big(-\sin(x)\big)(x^3 + 1)e^x - \cos(x)\,e^x(x^3 + 3x^2 + 1)}{(x^3 + 1)^2 (e^x)^2}.

Factor $e^x$ out of the numerator and cancel it against one factor of $(e^x)^2$ :

y' = \frac{e^x\big(-(x^3 + 1)\sin(x) - (x^3 + 3x^2 + 1)\cos(x)\big)}{(x^3 + 1)^2 e^x e^x} = \frac{-(x^3 + 1)\sin(x) - (x^3 + 3x^2 + 1)\cos(x)}{(x^3 + 1)^2 e^x}.

The product rule applied to $(uv)w$ , and again to $uv$ , gives the product rule for three functions.

Rule 1 (Product rule for three functions).

Let $u(x)$ , $v(x)$ , and $w(x)$ be differentiable functions. Then

(uvw)' = u'vw + uv'w + uvw'.

Example 5

Find the derivative:

y = x^2 e^x \sin(x).

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Name the three factors and their derivatives:

u = x^2, \quad u' = 2x, \qquad v = e^x, \quad v' = e^x, \qquad w = \sin(x), \quad w' = \cos(x).

Substitute the three factors and their derivatives into the product rule for three functions, and factor out $e^x$ :

y' = 2x e^x\sin(x) + x^2 e^x\sin(x) + x^2 e^x\cos(x) = e^x\big(2x\sin(x) + x^2\sin(x) + x^2\cos(x)\big).

Matching answer choices.

Substitute $u$ , $u'$ , $v$ , and $v'$ into the product rule or the quotient rule.
Simplify the result.
Compare the result with each choice.

Example 6

Determine which choice is the derivative of

f(x) = \frac{\ln(x)}{x^3 + x}.

(a) $\dfrac{\ln(x)(3x^2 + 1) - \frac{1}{x}(x^3 + x)}{(x^3 + x)^2}$	(b) $\dfrac{\frac{1}{x}(x^3 + x) + \ln(x)(3x^2 + 1)}{(x^3 + x)^2}$
(c) $\dfrac{\frac{1}{x}(x^3 + x) - \ln(x)(3x^2 + 1)}{x^3 + x}$	(d) $\dfrac{x^2 + 1 - (3x^2 + 1)\ln(x)}{(x^3 + x)^2}$

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Substitute $u = \ln(x)$ , $u' = 1/x$ , $v = x^3 + x$ , and $v' = 3x^2 + 1$ into the quotient rule:

f'(x) = \frac{\frac{1}{x}(x^3 + x) - \ln(x)(3x^2 + 1)}{(x^3 + x)^2}.

Simplify the first term of the numerator, $\frac{1}{x}(x^3 + x) = x^2 + 1$ :

f'(x) = \frac{x^2 + 1 - (3x^2 + 1)\ln(x)}{(x^3 + x)^2}.

Choice (d) matches this derivative. Choice (a) subtracts the numerator’s terms in the opposite order, choice (b) adds them, and choice (c) has no square in the denominator.

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2. Derivatives in R

Objective. Using R, obtain and plot the derivative of a function.

Plotting a derivative. The command D() from section 5 of the Lesson 29 Notes returns the derivative as a function. Give it a name, as in df = D(f(x) ~ x), so that later commands can use it. To plot df for $a \le x \le b$ , with the vertical axis from $c$ to $d$ , use the command

plotFun(df(x) ~ x, xlim = range(a, b), ylim = range(c, d))

Roots in R. For positive $x$ , the cube root of $x$ is x^(1/3), with the exponent in parentheses. Without them, R reads x^1/3 as $x^1$ divided by 3 and gives a wrong value with no error:

> 8^(1/3)
[1] 2
> 8^1/3
[1] 2.666667

Example 7

For the function

f(x) = \frac{x^5 - \sqrt{x}}{\cos(x)},

find $f'(1)$ in R. Then plot $f'$ for $0.2 \le x \le 1.2$ , with the vertical axis from $-5$ to $40$ .

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Define $f$ , differentiate it with D(), evaluate the derivative at 1, and plot it over the requested window. Used R.

f = makeFun((x^5 - sqrt(x)) / cos(x) ~ x)
df = D(f(x) ~ x)
df(1)
[1] 8.328671
plotFun(df(x) ~ x, xlim = range(0.2, 1.2), ylim = range(-5, 40))

To three decimals, $f'(1) \approx 8.329$ . As a check by the quotient rule with $u = x^5 - \sqrt{x}$ and $v = \cos(x)$ , the values $u(1) = 0$ , $u'(1) = 5 - \tfrac{1}{2} = 4.5$ , $v(1) = \cos(1)$ , and $v'(1) = -\sin(1)$ give the same value:

f'(1) = \frac{u'(1)v(1) - u(1)v'(1)}{v(1)^2} = \frac{4.5\cos(1) - 0}{\cos^2(1)} = \frac{4.5}{\cos(1)} \approx 8.329.

The plot is the graph of $f'$ over $0.2 \le x \le 1.2$ . The value $f'(1) \approx 8.329$ is the height of that graph at $x = 1$ , the marked point.

Example 8

For the function

g(x) = x\cos(x),

determine which curve is the graph of $g'(x)$ . Evaluate $g'(2)$ in R.

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Find $g'$ by the product rule, with $u = x$ and $v = \cos(x)$ :

g'(x) = (1)\cos(x) + x\big(-\sin(x)\big) = \cos(x) - x\sin(x).

Evaluate $g'(2)$ to locate one point on the graph of $g'$ . Used R.

g = makeFun(x*cos(x) ~ x)
dg = D(g(x) ~ x)
dg(2)
[1] -2.234742

The formula for $g'$ found by the product rule gives the same value:

g'(2) = \cos(2) - 2\sin(2) \approx -0.416 - 1.819 = -2.235.

At $x = 2$ , curve C is the only curve that passes through $(2, g'(2)) \approx (2, -2.235)$ . The other three curves sit above $-1$ there. Curve C is therefore the graph of $g'$ .

Example 9

The concentration of a drug in the blood, in milligrams per liter, $t$ hours after a dose, is modeled by

C(t) = \frac{20t}{t^2 + 4}.

Find $C'(1)$ and $C'(3)$ in R. Give the units of $C'$ . Determine whether the concentration is increasing or decreasing three hours after the dose.

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Define $C$ , differentiate it with D(), and evaluate $C'$ at 1 and 3. Used R.

C = makeFun(20*t / (t^2 + 4) ~ t)
dC = D(C(t) ~ t)
dC(1)
[1] 2.4
dC(3)
[1] -0.591716

The units of $C'$ are the units of $C$ per unit of $t$ : milligrams per liter per hour.

One hour after the dose, the concentration is increasing at 2.4 milligrams per liter per hour, because $C'(1) = 2.4$ is positive. Three hours after the dose, the concentration is decreasing at about 0.592 milligrams per liter per hour, because $C'(3) \approx -0.592$ is negative. On the graph of $C$ , the tangent line rises at $t = 1$ and falls at $t = 3$ .

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