Example 27: Log-differentiate $\sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}}$

Basics you need

Solution

1. Where is $y$ defined, and where can we take logs?

   Look at the sign of each factor.

   • $x^2 + 4 \ge 4 > 0$ for every $x$⁠.
   • $3x^2 + 4x + 5$⁠: here $a = 3 > 0$ and $b^2 - 4ac = 16 - 60 = -44 < 0$⁠, so it is positive for every $x$⁠. Completing the square shows it directly: $$\begin{aligned} &3x^2 + 4x + 5 \\ &= 3\left(x + \tfrac23\right)^2 + \tfrac{11}{3} > 0. \end{aligned}$$
   • $x - 3$ is positive for $x > 3$⁠, zero at $x = 3$⁠, negative for $x < 3$⁠.

   So the fraction under the root has the sign of $x - 3$⁠. A real square root needs it $\ge 0$⁠, so $y$ is defined for $x \ge 3$⁠. For the logarithms we need $x - 3 > 0$⁠, so we work on

   $$x > 3, \qquad\text{where } y > 0.$$
   The book stresses that in logarithmic differentiation the function and the quantities whose logs we take must be positive, "as otherwise their logarithms are not defined". Its solution does not say where this holds; here it is $x > 3$⁠. The point $x = 3$ (where $y = 0$⁠) is treated separately in Step 5.
   A quadratic with positive leading coefficient and negative discriminant is always positive (Basic 3).

2. Take logarithms and expand

   For $x > 3$⁠, write the root as a power,

   $$y = \left(\frac{(x-3)(x^2+4)}{3x^2+4x+5}\right)^{1/2},$$

   and take $\log$ of both sides:

   $$\begin{aligned} \log y = \tfrac12\big[&\log(x-3) + \log(x^2+4) \\ &- \log(3x^2+4x+5)\big]. \end{aligned}$$
   Three log laws are used, each valid because every quantity is positive for $x > 3$⁠: $\log p^{1/2} = \frac12\log p$ brings the root down as a factor; $\log\frac{AB}{C} = \log A + \log B - \log C$ splits the product and the quotient. The right side is now a sum of simple logs, each of which we can differentiate with one chain-rule step.
   $\log_b pq = \log_b p + \log_b q$⁠, $\log_b\frac{x}{y} = \log_b x - \log_b y$⁠, $\log_b p^n = n\log_b p$ (Section 5.4). Basics: exponentials and logarithms

3. Differentiate both sides with respect to $x$

   Left side: the outer function is $\log$⁠, the inner is $y$⁠, so $\frac{d}{dx}\log y = \frac1y\cdot\frac{dy}{dx}$⁠. Right side: use $\frac{d}{dx}\log g = \frac{g'}{g}$ on each term:

   $$\begin{gathered} \frac{d}{dx}\log(x-3) = \frac{1}{x-3}, \\ \frac{d}{dx}\log(x^2+4) = \frac{2x}{x^2+4}, \end{gathered}$$ $$\frac{d}{dx}\log(3x^2+4x+5) = \frac{6x+4}{3x^2+4x+5}.$$

   So

   $$\begin{aligned} \frac1y\cdot\frac{dy}{dx} = \frac12\Bigg[&\frac{1}{x-3} + \frac{2x}{x^2+4} \\ &- \frac{6x+4}{3x^2+4x+5}\Bigg]. \end{aligned}$$
   Each log term is a composite with outer $\log$ and an inner polynomial $g$⁠. The chain rule gives $\frac1g$ times $g'$⁠. The inner derivatives are $1$⁠, $2x$ and $6x + 4$ by the power rule. Forgetting them (writing just $\frac1g$⁠) is the classic error.
   Theorem 5 (2) and the chain rule: $\frac{d}{dx}\log g(x) = \frac{g'(x)}{g(x)}$⁠. Lesson: logarithmic differentiation

4. Multiply by $y$ and substitute it back

   $$\begin{aligned} \frac{dy}{dx} = \frac{y}{2}\Bigg[&\frac{1}{x-3} + \frac{2x}{x^2+4} \\ &- \frac{6x+4}{3x^2+4x+5}\Bigg], \end{aligned}$$

   that is, for $x > 3$⁠,

   $$\begin{aligned} \frac{dy}{dx} = \ &\frac12\sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}} \\ &\times\Bigg[\frac{1}{x-3} + \frac{2x}{x^2+4} \\ &\qquad - \frac{6x+4}{3x^2+4x+5}\Bigg]. \end{aligned}$$
   The left side was $\frac1y\frac{dy}{dx}$⁠; multiplying both sides by $y$ isolates $\frac{dy}{dx}$⁠. We then replace $y$ by its formula so the answer is in $x$ alone. This is exactly the book's final form.
   Rearranging: if $\frac1y\,Y = R$ with $y \ne 0$⁠, then $Y = yR$⁠. Basics: solving for dy/dx

5. What happens at $x = 3$⁠?

   At $x = 3$⁠, $y = 0$ and the log method cannot be used. Near $x = 3$⁠, $y \approx \sqrt{\frac{13}{44}(x - 3)}$ (using $3^2 + 4 = 13$ and $27 + 12 + 5 = 44$⁠), so the right-hand difference quotient behaves like

   $$\frac{y(3+h) - 0}{h} \approx \sqrt{\frac{13}{44}}\cdot\frac{1}{\sqrt h},$$

   which grows without bound as $h \to 0^+$⁠. The graph starts at $(3, 0)$ with a vertical tangent, and $y$ has no derivative there.

   This explains why the formula is valid only for $x > 3$⁠: the factor $\frac{1}{x-3}$ in the bracket blows up as $x \to 3$⁠, and the derivative really does not exist at $3$⁠. The function is not defined at all for $x < 3$⁠.
   A derivative exists only when the limit of the difference quotient is finite (Section 5.3). Basics: slope at a point

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