Chapter 5 · Continuity and Differentiability

Exercise 5.5 · Q16: $f'(1)$ for $(1+x)(1+x^2)(1+x^4)(1+x^8)$

Exercise 5.5Logarithmic differentiation · Section 5.5

The question

Find the derivative of the function given by

f(x) = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8)

and hence find $f'(1)$ ⁠.

Exam answer

Let $y = f(x)$ ⁠. Taking log on both sides,

\begin{aligned} \log y &= \log(1 + x) + \log(1 + x^2) \\ &\quad + \log(1 + x^4) \\ &\quad + \log(1 + x^8) \end{aligned}

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

\begin{aligned} \frac1y\frac{dy}{dx} &= \frac{1}{1 + x} + \frac{2x}{1 + x^2} \\ &\quad + \frac{4x^3}{1 + x^4} + \frac{8x^7}{1 + x^8} \end{aligned}

\begin{aligned} \therefore\ f'(x) &= (1 + x)(1 + x^2) \\ &\quad\times(1 + x^4)(1 + x^8) \\ &\quad\times\bigg[\frac{1}{1 + x} + \frac{2x}{1 + x^2} \\ &\qquad + \frac{4x^3}{1 + x^4} + \frac{8x^7}{1 + x^8}\bigg] \end{aligned}

for $x \ne -1$ ⁠. Putting $x = 1$ ⁠,

\begin{aligned} f'(1) &= 2\cdot 2\cdot 2\cdot 2 \\ &\quad\times\left[\frac12 + \frac22 + \frac42 + \frac82\right] \\ &= 16\times\frac{15}{2} \end{aligned}

$\therefore\ f'(1) = 120$

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

The one idea this problem is built on

A product of four factors is tedious with the product rule. Taking logs turns the product into a sum of four logs, each differentiated in one line. The logs need every factor positive, so we find where that is true, and then see that the final formula holds even beyond that.

Given, to find, and the plan

Given: $f(x) = (1 + x)(1 + x^2)$ $\times\,(1 + x^4)(1 + x^8)$ ⁠, a polynomial, defined for all real $x$ ⁠.
To find: $f'(x)$ ⁠, and $f'(1)$ ⁠.
Plan: (1) Find where $f > 0$ ⁠. (2) Take logs. (3) Differentiate. (4) Multiply by $f$ ⁠. (5) Extend to $x < -1$ and look at $x = -1$ ⁠. (6) Put $x = 1$ ⁠. (7) Cross-check with a closed form for $f$ ⁠.

Basics you need

1Log of a product

$\log(pq) = \log p + \log q$ for $p, q > 0$ (Section 5.4); it extends to any number of positive factors. Basics: exponentials and logarithms

2Derivative of $\log g(x)$

For $g > 0$ ⁠: $\frac{d}{dx}\log g = \frac{g'}{g}$ (Theorem 5 with the chain rule). Basics: derivative rules

3Geometric sums

$1 + x + x^2 + \cdots + x^{15} = \frac{1 - x^{16}}{1 - x}$ for $x \ne 1$ ⁠, and $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$ ⁠. Basics: solving for dy/dx

Solution

Where is $f(x)$ positive?

$1 + x^2$ ⁠, $1 + x^4$ ⁠, $1 + x^8$ are all $\ge 1 > 0$ for every $x$ ⁠, because even powers are never negative. The factor $1 + x$ is positive exactly when $x > -1$ ⁠. So
$f(x) > 0 \quad\text{for } x > -1.$
Logarithmic differentiation needs the logged quantities positive. Here that means $x > -1$ at first; Step 5 removes this restriction.

Positivity requirement (Section 5.5). Lesson: logarithmic differentiation
Take logarithms

For $x > -1$ ⁠, every factor is positive, so
$\begin{aligned} \log f(x) = \ &\log(1+x) + \log(1+x^2) \\ &+ \log(1+x^4) + \log(1+x^8). \end{aligned}$
The log of a product of positive numbers is the sum of their logs. Sums are differentiated term by term.

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

Each term is $\log g$ with $g$ a simple polynomial, so it differentiates to $\frac{g'}{g}$ ⁠:
$\begin{aligned} \frac{f'(x)}{f(x)} = \ &\frac{1}{1+x} + \frac{2x}{1+x^2} \\ &+ \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8}. \end{aligned}$
The inner derivatives are $(1 + x)' = 1$ ⁠, $(1 + x^2)' = 2x$ ⁠, $(1 + x^4)' = 4x^3$ ⁠, $(1 + x^8)' = 8x^7$ by the power rule. On the left, $\frac{d}{dx}\log f = \frac{f'}{f}$ ⁠.

Chain rule (Theorem 4, Section 5.3.1); $(\log x)' = \frac1x$ (Theorem 5); $(x^n)' = nx^{n-1}$ (Table 5.3). Basics: derivative rules
Multiply by $f(x)$
$\begin{aligned} f'(x) = \ &(1+x)(1+x^2)(1+x^4)(1+x^8) \\ &\times\Bigg[\frac{1}{1+x} + \frac{2x}{1+x^2} \\ &\qquad + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8}\Bigg]. \end{aligned}$
Multiplying $\frac{f'}{f}$ by $f$ leaves $f'$ ⁠. If we multiply out, each fraction cancels its own factor and leaves the product of the other three: that is exactly what the product rule for four factors would give.

Logarithmic differentiation, $f' = f\cdot(\log f)'$ ⁠. Lesson: logarithmic differentiation
What about $x < -1$ and $x = -1$ ⁠?

For $x < -1$ ⁠, $f(x) < 0$ ⁠, so use
$-f(x) = (-1-x)(1+x^2)(1+x^4)(1+x^8) > 0.$
Taking logs of $-f$ gives the same sum, except that the first term is $\log(-1-x)$ ⁠, whose derivative is $\frac{-1}{-1-x} = \frac{1}{1+x}$ ⁠. Also $\frac{d}{dx}\log(-f) = \frac{-f'}{-f} = \frac{f'}{f}$ ⁠. So the formula of Step 4 holds for every $x \ne -1$ ⁠.

At $x = -1$ the bracket is undefined, but $f$ is a polynomial and the product rule gives directly: only the term where $(1+x)$ is differentiated survives, so
$f'(-1) = 1\cdot(1+1)(1+1)(1+1) = 8.$
The book's answer does not mention the restriction. After multiplying out (Step 4), the expression is a polynomial and is valid everywhere, including $x = -1$ ⁠; written with the bracket, it needs $x \ne -1$ ⁠.

Product rule (Section 5.3). Basics: derivative rules
Put $x = 1$

$f(1) = 2\cdot 2\cdot 2\cdot 2 = 16$ ⁠, and the bracket is
$\frac12 + \frac22 + \frac42 + \frac82 = \frac{15}{2}.$
So
$f'(1) = 16\times\frac{15}{2} = 120.$
At $x = 1$ every power of $x$ is $1$ ⁠, so each denominator is $2$ and the numerators are $1, 2, 4, 8$ ⁠, adding to $15$ ⁠.

Substitution into the formula of Step 4. Basics: derivative rules
Cross-check with a closed form

Multiply $f(x)$ by $1 - x$ and use $(1 - a)(1 + a) = 1 - a^2$ four times:
$\begin{aligned} (1 - x)f(x) &= (1 - x^2)(1+x^2)(1+x^4)(1+x^8) \\ &= (1 - x^4)(1+x^4)(1+x^8) \\ &= (1 - x^8)(1+x^8) = 1 - x^{16}. \end{aligned}$
So for $x \ne 1$ ⁠,
$\begin{aligned} f(x) &= \frac{1 - x^{16}}{1 - x} \\ &= 1 + x + x^2 + \cdots + x^{15}. \end{aligned}$
At $x = 1$ both sides equal $16$ ( $f(1) = 2^4$ ⁠, and the sum has $16$ terms, each $1$ ⁠), so $f(x) = 1 + x + \cdots + x^{15}$ for every $x$ ⁠. Then
$\begin{aligned} f'(x) &= 1 + 2x + 3x^2 + \cdots + 15x^{14}, \\ f'(1) &= 1 + 2 + \cdots + 15 = \frac{15\times 16}{2} = 120. \end{aligned}$
The same value by a completely different route. It also confirms Step 5: $f'(-1) = 1 - 2 + 3 - \cdots + 15$ ⁠, which is seven pairs $(1 - 2), (3 - 4), \ldots, (13 - 14)$ ⁠, each $-1$ ⁠, plus $15$ ⁠: $-7 + 15 = 8$ ⁠.

Difference of squares; geometric sum; sum of the first $n$ natural numbers (Class XI). Basics: solving for dy/dx

Result

\begin{aligned} f'(x) = \ &(1+x)(1+x^2)(1+x^4)(1+x^8) \\ &\times\Bigg[\frac{1}{1+x} + \frac{2x}{1+x^2} \\ &\qquad + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8}\Bigg] \end{aligned}

for $x \ne -1$ (and $f'(-1) = 8$ ⁠); $f'(1) = 120$ ⁠. (Textbook answer: the same expression and $f'(1) = 120$ ⁠.)

Check it with numbers

The central difference $\frac{f(1.00001) - f(0.99999)}{0.00002}$ gives $120.00000$ to five decimal places. At $x = 0.5$ ⁠: $f(0.5) = 1.5\times 1.25\times 1.0625$ $\times\,1.00390625 = 1.9999695$ ⁠, the bracket is $1.9995117$ ⁠, so $f'(0.5) = 3.9989624$ ⁠; and the sum $1 + 2(0.5) + 3(0.5)^2 + \cdots$ up to $15(0.5)^{14}$ is $3.9989624$ as well.

Common mistakes

Dropping inner derivatives: $\frac{d}{dx}\log(1 + x^4) = \frac{4x^3}{1+x^4}$ ⁠, not $\frac{1}{1+x^4}$ ⁠.
Forgetting to multiply by $f(x)$ at the end, so that $f'(1)$ comes out as $\frac{15}{2}$ instead of $120$ ⁠.
Using $f(1) = 2 + 2 + 2 + 2$ : it is a product, $2^4 = 16$ ⁠.
Ignoring where the logs exist. The log step needs $x > -1$ (extendable to $x \ne -1$ ⁠).

What to take away

For a long product, $\frac{f'}{f}$ is the sum of $\frac{g'}{g}$ over the factors $g$ ⁠.
The same formula works where $f < 0$ by logging $-f$ (or $|f|$ ⁠); it fails only where a factor is zero.
A neat closed form (here $\frac{1 - x^{16}}{1 - x}$ ⁠) is a valuable independent check.

Every fact used, and where it comes from

Fact	Used in	Source
Positivity requirement for logarithmic differentiation	Steps 1, 5	Section 5.5 (Lesson: logarithmic differentiation)
$\log pq = \log p + \log q$	Steps 2, 5	Section 5.4
$\frac{d}{dx}\log x = \frac1x$ ⁠; chain rule	Steps 3, 5	Theorem 5, Section 5.4; Theorem 4, Section 5.3.1
Power rule; product rule	Steps 3, 5, 7	Table 5.3; algebra of derivatives, Section 5.3
Geometric sum; $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$	Step 7	Sequences and series (Class XI)

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Exercise 5.5 · Q16: f′(1)f'(1)f′(1) for (1+x)(1+x2)(1+x4)(1+x8)(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x)(1+x2)(1+x4)(1+x8)

Basics you need

Solution

Where is f(x)f(x)f(x) positive?

Take logarithms

Differentiate both sides

Multiply by f(x)f(x)f(x)

What about x<−1x < -1x<−1 and x=−1x = -1x=−1⁠?

Put x=1x = 1x=1

Cross-check with a closed form

Every fact used, and where it comes from

Exercise 5.5 · Q16: $f'(1)$ for $(1+x)(1+x^2)(1+x^4)(1+x^8)$

Where is $f(x)$ positive?

Multiply by $f(x)$

What about $x < -1$ and $x = -1$ ⁠?

Put $x = 1$