Exercise 5.5 · Q2: Differentiate $\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$

Basics you need

Solution

1. Where is $y$ defined, and where is it positive?

   Each factor $x - a$ changes sign only at $x = a$⁠. The points $1, 2, 3, 4, 5$ split the line into six intervals; count the negative factors in each:

   Factor	$x<1$	$(1,2)$	$(2,3)$	$(3,4)$	$(4,5)$	$x>5$
   $x-1$	−	+	+	+	+	+
   $x-2$	−	−	+	+	+	+
   $x-3$	−	−	−	+	+	+
   $x-4$	−	−	−	−	+	+
   $x-5$	−	−	−	−	−	+
   $Q(x)$	−	+	−	+	−	+

   A real square root needs $Q(x) \ge 0$⁠, and $Q$ is undefined at $3, 4, 5$⁠. $Q = 0$ exactly at $x = 1$ and $x = 2$⁠. So

   $$y \text{ is defined on } [1, 2]\,\cup\,(3, 4)\,\cup\,(5, \infty),$$ $$y > 0 \text{ on } (1, 2)\,\cup\,(3, 4)\,\cup\,(5, \infty).$$
   Imitating Example 27 one would write $\log(x-1)$⁠, $\log(x-2)$ and so on, which all exist only for $x > 5$⁠. On $(1, 2)$ and $(3, 4)$ the fraction is positive although some factors are negative, so the function is defined there too. The sign table finds all three pieces; the next step makes one formula work on all of them.
   Sign of a product or quotient from the number of negative factors (Basic 3). Basics: functions

2. Take logarithms, with absolute values

   On $(1,2)$⁠, $(3,4)$ and $(5,\infty)$⁠, $Q > 0$⁠, so $Q = |Q|$ and $\log y = \frac12\log Q$⁠. Since $|Q|$ is a product and quotient of the positive numbers $|x - a|$⁠,

   $$\begin{aligned} \log y = \tfrac12\big[&\log|x-1| + \log|x-2| \\ &- \log|x-3| - \log|x-4| \\ &- \log|x-5|\big]. \end{aligned}$$
   $|Q| = \frac{|x-1|\,|x-2|}{|x-3|\,|x-4|\,|x-5|}$ because the absolute value of a product or quotient is the product or quotient of the absolute values. Each $|x - a|$ is positive (no factor is zero on these intervals), so the log laws apply. For $x > 5$ every bar can be dropped, giving the line Example 27 would suggest.
   $\log\sqrt{p} = \frac12\log p$⁠, $\log\frac{pq}{rst} = \log p + \log q - \log r - \log s - \log t$ (Section 5.4). Basics: exponentials and logarithms

3. Differentiate both sides

   Left side (chain rule, outer $\log$⁠, inner $y$⁠): $\frac1y\frac{dy}{dx}$⁠. Right side: $\frac{d}{dx}\log|x - a| = \frac{1}{x-a}$ for each term (Basic 2). So

   $$\begin{aligned} \frac1y\cdot\frac{dy}{dx} = \frac12\Bigg[&\frac{1}{x-1} + \frac{1}{x-2} \\ &- \frac{1}{x-3} - \frac{1}{x-4} \\ &- \frac{1}{x-5}\Bigg]. \end{aligned}$$
   Each inner function $x - a$ has derivative $1$⁠, so each term is simply $\frac{1}{x-a}$⁠. The minus signs of the denominator's factors carry through unchanged.
   $\frac{d}{dx}\log|g| = \frac{g'}{g}$ with $g = x - a$⁠, $g' = 1$⁠. Lesson: logarithmic differentiation

4. Multiply by $y$

   For $x$ in $(1,2)$⁠, $(3,4)$ or $(5,\infty)$⁠:

   $$\begin{aligned} \frac{dy}{dx} = \ &\frac12\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \\ &\times\Bigg[\frac{1}{x-1} + \frac{1}{x-2} \\ &\qquad - \frac{1}{x-3} - \frac{1}{x-4} \\ &\qquad - \frac{1}{x-5}\Bigg]. \end{aligned}$$
   Multiplying by $y$ (positive here) isolates $\frac{dy}{dx}$⁠, and substituting $y$ back gives the answer in $x$ alone. This is the textbook's answer, now with the set where it holds.
   Rearranging: $\frac1y Y = R$ with $y \ne 0$ gives $Y = yR$⁠. Basics: solving for dy/dx

5. What happens at $x = 1$ and $x = 2$⁠?

   There $y = 0$⁠, so the log method cannot be used, and $y$ exists only on one side of each point (inside $[1, 2]$⁠). At $x = 1$⁠, take $0 < h < 1$⁠. Then

   $$\begin{aligned} Q(1+h) &= \frac{h\,(h-1)}{(h-2)(h-3)(h-4)} \\ &= \frac{h\,(1-h)}{(2-h)(3-h)(4-h)} \end{aligned}$$

   (numerator and denominator both changed sign), so the right-hand difference quotient is

   $$\begin{aligned} \frac{y(1+h) - 0}{h} &= \sqrt{\frac{Q(1+h)}{h^2}} \\ &= \frac{1}{\sqrt h}\sqrt{\frac{1-h}{(2-h)(3-h)(4-h)}}. \end{aligned}$$

   As $h \to 0^+$ the second root tends to $\sqrt{1/24}$⁠, while $\frac{1}{\sqrt h}$ grows without bound, so the quotient tends to $+\infty$⁠. At $x = 2$⁠, in the same way,

   $$Q(2-h) = \frac{h\,(1-h)}{(1+h)(2+h)(3+h)},$$

   and the left-hand quotient $\frac{y(2-h) - 0}{-h}$ equals $-\frac{1}{\sqrt h}$ times a root that tends to $\sqrt{1/6}$⁠, so it tends to $-\infty$⁠.

   So the graph meets the $x$-axis at $1$ and $2$ with vertical tangents, and $y$ has no derivative there. The points $3, 4, 5$ are not in the domain at all, so nothing is asked there. Hence the formula of Step 4 holds exactly on $(1,2)$⁠, $(3,4)$ and $(5,\infty)$⁠.
   A derivative (or one-sided derivative) is a finite limit of difference quotients (Section 5.3). Basics: slope at a point

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