Chapter 5 · Continuity and Differentiability

Exercise 5.5 · Q5: Differentiate $(x+3)^2(x+4)^3(x+5)^4$

Exercise 5.5Logarithmic differentiation · Section 5.5

The question

Differentiate the function w.r.t. $x$ ⁠:

(x+3)^2\,.\,(x+4)^3\,.\,(x+5)^4

Exam answer

Let $y = (x + 3)^2(x + 4)^3(x + 5)^4$ ⁠. Taking log on both sides (where $y \ne 0$ ⁠),

\begin{aligned} \log|y| &= 2\log|x + 3| \\ &\quad + 3\log|x + 4| \\ &\quad + 4\log|x + 5| \end{aligned}

Differentiating both sides w.r.t. $x$ ⁠,

\frac1y\cdot\frac{dy}{dx} = \frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5}

Multiplying by $y$ ⁠,

\begin{aligned} \frac{dy}{dx} &= (x + 3)^2(x + 4)^3(x + 5)^4 \\ &\quad\times\bigg[\frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5}\bigg] \\ &= (x + 3)(x + 4)^2(x + 5)^3 \\ &\quad\times\big[2(x + 4)(x + 5) \\ &\qquad + 3(x + 3)(x + 5) \\ &\qquad + 4(x + 3)(x + 4)\big] \end{aligned}

Expanding the bracket, $2(x^2 + 9x + 20)$ $+ 3(x^2 + 8x + 15)$ $+ 4(x^2 + 7x + 12)$ $= 9x^2 + 70x + 133$ ⁠.

$\begin{aligned} \therefore\ \dfrac{dy}{dx} &= (x + 3)(x + 4)^2(x + 5)^3 \\ &\quad\times(9x^2 + 70x + 133) \end{aligned}$

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Logarithms turn a product of powers into a sum of multiples: $\log|y|$ equals $2\log|x+3| + 3\log|x+4|$ $+\ 4\log|x+5|$ ⁠. Each term differentiates in one line. Multiplying back by $y$ and taking out the common factors gives a compact answer. The absolute values are needed because $y$ is negative for some $x$ ⁠; and since $y$ is a polynomial, the final answer turns out to hold for every $x$ ⁠, including the zeros where the log method itself cannot be used.

Given, to find, and the plan

Given: $y = (x+3)^2(x+4)^3(x+5)^4$ ⁠, a polynomial.
To find: $\frac{dy}{dx}$ in factored form.
Plan: (1) Where is $y \ne 0$ ⁠? (2) Take $\log|y|$ ⁠. (3) Differentiate. (4) Multiply by $y$ and factor. (5) Expand the remaining bracket. (6) Confirm with the product rule, which also covers $x = -3, -4, -5$ ⁠.

Basics you need

1Log laws

$\log(pq) = \log p + \log q$ and $\log p^n = n\log p$ for $p, q > 0$ (Section 5.4). With $|a^n| = |a|^n$ ⁠: $\log|a^n| = n\log|a|$ for $a \ne 0$ ⁠. Basics: exponentials and logarithms

2 $\frac{d}{dx}\log|g| = \frac{g'}{g}$ where $g \ne 0$

For $g > 0$ this is the chain rule with Theorem 5; for $g < 0$ ⁠, $\frac{d}{dx}\log(-g) = \frac{-g'}{-g} = \frac{g'}{g}$ ⁠. So $\frac{d}{dx}\log|x + a| = \frac{1}{x + a}$ ⁠. Basics: derivative rules

3Product rule and power rule

$(uvw)' = u'vw + uv'w + uvw'$ (the product rule used twice, Section 5.3); $\frac{d}{dx}(x+a)^n = n(x+a)^{n-1}$ (Table 5.3 with the chain rule). Basics: derivative rules

Solution

Where is $y$ non-zero, and what sign does it have?

$y = 0$ exactly at $x = -3, -4, -5$ ⁠. Elsewhere $y \ne 0$ ⁠, but it is not always positive: $(x+3)^2$ and $(x+5)^4$ are never negative, while $(x+4)^3$ has the sign of $x + 4$ ⁠. So $y < 0$ for $x < -4$ (except $x = -5$ ⁠) and $y > 0$ for $x > -4$ (except $x = -3$ ⁠). For example, at $x = -4.5$ ⁠:
$y = (-1.5)^2(-0.5)^3(0.5)^4 = -0.017578125.$
The book's method takes $\log y$ ⁠, which needs $y > 0$ ⁠. Since $y$ is negative on part of the line, we take $\log|y|$ ⁠, which exists wherever $y \ne 0$ ⁠, that is, for $x \ne -3, -4, -5$ ⁠.

An even power is never negative; an odd power keeps the sign of its base. Basics: functions
Take $\log|y|$

For $x \ne -3, -4, -5$ ⁠:
$\begin{aligned} \log|y| &= \log\big(|x+3|^2\,|x+4|^3\,|x+5|^4\big) \\ &= 2\log|x+3| + 3\log|x+4| \\ &\quad + 4\log|x+5|. \end{aligned}$
$|y| = |x+3|^2|x+4|^3|x+5|^4$ because the absolute value of a product of powers is the product of the powers of the absolute values. All three $|x + a|$ are positive here, so the product law and the power law apply. The exponents come down as multipliers.

$\log pq = \log p + \log q$ ⁠, $\log p^n = n\log p$ (Section 5.4). Basics: exponentials and logarithms
Differentiate both sides

Left: $\frac{d}{dx}\log|y| = \frac1y\frac{dy}{dx}$ ⁠. Right: $\frac{d}{dx}\log|x + a| = \frac{1}{x+a}$ (Basic 2). So
$\frac1y\cdot\frac{dy}{dx} = \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5}.$
Each inner function $x + a$ has derivative $1$ ⁠, so each log term gives $\frac{1}{x+a}$ times its multiplier. The sum on the right is the "relative rate of change" of $y$ ⁠.

$\frac{d}{dx}\log|g| = \frac{g'}{g}$ ⁠. Lesson: logarithmic differentiation
Multiply by $y$ and take out common factors

Multiplying each fraction by $y = (x+3)^2(x+4)^3(x+5)^4$ cancels one factor:
$\begin{aligned} \frac{dy}{dx} &= 2(x+3)(x+4)^3(x+5)^4 \\ &\quad + 3(x+3)^2(x+4)^2(x+5)^4 \\ &\quad + 4(x+3)^2(x+4)^3(x+5)^3. \end{aligned}$
Every term contains $(x+3)(x+4)^2(x+5)^3$ ⁠. Taking it out:
$\begin{aligned} \frac{dy}{dx} = \ &(x+3)(x+4)^2(x+5)^3 \\ &\times\big[2(x+4)(x+5) \\ &\quad + 3(x+3)(x+5) \\ &\quad + 4(x+3)(x+4)\big]. \end{aligned}$
For instance $y\cdot\frac{2}{x+3} = 2(x+3)(x+4)^3(x+5)^4$ ⁠: one power of $x + 3$ cancels. The common factor uses the lowest power of each bracket that appears in all three terms: $(x+3)^1$ ⁠, $(x+4)^2$ ⁠, $(x+5)^3$ ⁠.

Factorising out a common factor. Basics: solving for dy/dx
Expand the bracket
$\begin{aligned} 2(x+4)(x+5) &= 2x^2 + 18x + 40, \\ 3(x+3)(x+5) &= 3x^2 + 24x + 45, \\ 4(x+3)(x+4) &= 4x^2 + 28x + 48. \end{aligned}$
Adding: $9x^2 + 70x + 133$ ⁠. Therefore
$\frac{dy}{dx} = (x+3)(x+4)^2(x+5)^3(9x^2 + 70x + 133).$
$(x+4)(x+5) = x^2 + 9x + 20$ ⁠, $(x+3)(x+5) = x^2 + 8x + 15$ ⁠, $(x+3)(x+4) = x^2 + 7x + 12$ ⁠. Collecting: $x^2$ terms $2 + 3 + 4 = 9$ ⁠; $x$ terms $18 + 24 + 28 = 70$ ⁠; constants $40 + 45 + 48 = 133$ ⁠. This is the textbook's answer.

Multiplying out brackets and collecting like terms. Basics: solving for dy/dx
The answer holds at $x = -3, -4, -5$ too

The log method needed $y \ne 0$ ⁠. But the first display of Step 4 is exactly what the product rule gives directly, with $u = (x+3)^2$ ⁠, $v = (x+4)^3$ ⁠, $w = (x+5)^4$ ⁠, $u' = 2(x+3)$ ⁠, $v' = 3(x+4)^2$ ⁠, $w' = 4(x+5)^3$ ⁠:
$\begin{aligned} u'vw &= 2(x+3)(x+4)^3(x+5)^4, \\ uv'w &= 3(x+3)^2(x+4)^2(x+5)^4, \\ uvw' &= 4(x+3)^2(x+4)^3(x+5)^3. \end{aligned}$
The product rule has no sign or non-zero condition, so the factored answer of Step 5 is valid for every real $x$ ⁠.

For a product of powers, logarithmic differentiation is a quick way to organise the product rule, and the result can then be checked to hold everywhere. For example, at $x = -3$ the answer gives $\frac{dy}{dx} = 0$ ⁠: the graph touches the $x$ -axis there, because of the squared factor $(x+3)^2$ ⁠.

Product rule used twice; chain rule for $(x + a)^n$ ⁠. Basics: derivative rules

Result

For every real $x$ ⁠,

\frac{dy}{dx} = (x+3)(x+4)^2(x+5)^3(9x^2 + 70x + 133).

(Textbook answer: $(x+3)(x+4)^2(x+5)^3$ $(9x^2 + 70x + 133)$ ⁠, the same.)

Check it with numbers

At $x = 0$ ⁠: $y = 9\times 64\times 625 = 360000$ ⁠, and Step 3 gives $\frac1y\frac{dy}{dx} = \frac23 + \frac34 + \frac45 = \frac{133}{60}$ ⁠, so $\frac{dy}{dx} = 360000\times\frac{133}{60} = 798000$ ⁠. The factored answer gives $3\times 16\times 125\times 133 = 798000$ ⁠. The central difference quotient with $h = 0.0001$ gives $798000.00$ (to two decimals).

At $x = -4.5$ ⁠, where $y < 0$ ⁠: $\frac{2}{-1.5} + \frac{3}{-0.5} + \frac{4}{0.5} = \frac23$ ⁠, so $\frac{dy}{dx} = -0.017578125\times\frac23$ $= -0.01171875$ ⁠; the factored answer gives $(-1.5)(0.25)(0.125)(0.25)$ $= -0.01171875$ ⁠. The $\log|y|$ route is right where $y$ is negative.

Common mistakes

Forgetting the exponents as multipliers: $\log(x+4)^3 = 3\log(x+4)$ ⁠, giving $\frac{3}{x+4}$ ⁠, not $\frac{1}{x+4}$ ⁠.
Factoring out the wrong powers. Take the lowest power of each bracket present in all terms: $(x+3)^1(x+4)^2(x+5)^3$ ⁠.
Arithmetic in the bracket: check $9x^2 + 70x + 133$ at one value, e.g. $x = 0$ ⁠: $40 + 45 + 48 = 133$ ⁠.
Claiming the answer fails at $-3, -4, -5$ ⁠. Only the log method does; the result itself holds everywhere (Step 6).

What to take away

For $y = (x-a)^m(x-b)^n\cdots$ ⁠: $\frac{y'}{y} = \frac{m}{x-a} + \frac{n}{x-b} + \cdots$ ⁠.
Use $\log|y|$ when $y$ can be negative; the derivative formula is unchanged.
Factor the answer by the lowest common powers; then expand only the small bracket left over.

Every fact used, and where it comes from

Fact	Used in	Source
Logarithmic differentiation needs a positive function	Step 1	Section 5.5 (Lesson: logarithmic differentiation)
$\log pq = \log p + \log q$ ⁠, $\log p^n = n\log p$	Step 2	Section 5.4
$\frac{d}{dx}\log\|g\| = \frac{g'}{g}$ ⁠, $g \ne 0$	Step 3	Basic 2 (from Theorem 4, Section 5.3.1, and Theorem 5, Section 5.4)
Product rule; $\frac{d}{dx}x^n = nx^{n-1}$	Step 6	Algebra of derivatives (2), Section 5.3; Table 5.3
Chain rule	Steps 3, 6	Theorem 4, Section 5.3.1

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Exercise 5.5 · Q5: Differentiate (x+3)2(x+4)3(x+5)4(x+3)^2(x+4)^3(x+5)^4(x+3)2(x+4)3(x+5)4

Basics you need

Solution

Where is yyy non-zero, and what sign does it have?

Take log⁡∣y∣\log|y|log∣y∣

Differentiate both sides

Multiply by yyy and take out common factors

Expand the bracket

The answer holds at x=−3,−4,−5x = -3, -4, -5x=−3,−4,−5 too

Every fact used, and where it comes from

Exercise 5.5 · Q5: Differentiate $(x+3)^2(x+4)^3(x+5)^4$

Where is $y$ non-zero, and what sign does it have?

Take $\log|y|$

Multiply by $y$ and take out common factors

The answer holds at $x = -3, -4, -5$ too