Exercise 5.5 · Q17: Differentiate $(x^2-5x+8)(x^3+7x+9)$ three ways

Basics you need

Part (i): by the product rule

1. Differentiate each factor

   $$\begin{aligned} u &= x^2 - 5x + 8, & u' &= 2x - 5, \\ v &= x^3 + 7x + 9, & v' &= 3x^2 + 7. \end{aligned}$$
   Each factor is a polynomial; differentiate term by term with the power rule. Constants ($8$ and $9$⁠) have derivative $0$⁠.
   Power rule (Table 5.3); sum rule (Section 5.3). Basics: derivative rules

2. Apply the product rule

   $$\begin{aligned} \frac{dy}{dx} &= u'v + uv' \\ &= (2x - 5)(x^3 + 7x + 9) \\ &\quad + (x^2 - 5x + 8)(3x^2 + 7). \end{aligned}$$
   The product rule: derivative of the first times the second, plus the first times the derivative of the second. This is already a correct answer; we expand it so that the three methods can be compared.
   Product (Leibnitz) rule (Section 5.3). Basics: derivative rules

3. Expand the two products

   $$\begin{aligned} &(2x - 5)(x^3 + 7x + 9) \\ &= 2x^4 + 14x^2 + 18x - 5x^3 - 35x - 45 \\ &= 2x^4 - 5x^3 + 14x^2 - 17x - 45, \end{aligned}$$ $$\begin{aligned} &(x^2 - 5x + 8)(3x^2 + 7) \\ &= 3x^4 + 7x^2 - 15x^3 - 35x + 24x^2 + 56 \\ &= 3x^4 - 15x^3 + 31x^2 - 35x + 56. \end{aligned}$$
   Multiply every term of the first bracket by every term of the second, then collect like powers: $18x - 35x = -17x$ and $7x^2 + 24x^2 = 31x^2$⁠.
   Distributive law; collecting like terms. Basics: solving for dy/dx

4. Add

   $$\begin{aligned} \frac{dy}{dx} &= (2 + 3)x^4 + (-5 - 15)x^3 + (14 + 31)x^2 \\ &\quad + (-17 - 35)x + (-45 + 56) \\ &= 5x^4 - 20x^3 + 45x^2 - 52x + 11. \end{aligned}$$
   Add the coefficients of each power of $x$ separately.
   Collecting like terms. Basics: solving for dy/dx

Part (ii): by expanding first

1. Multiply out the product

   Multiply $v = x^3 + 7x + 9$ by each term of $u$⁠:

   $$\begin{aligned} x^2\cdot v &= x^5 + 7x^3 + 9x^2, \\ -5x\cdot v &= -5x^4 - 35x^2 - 45x, \\ 8\cdot v &= 8x^3 + 56x + 72. \end{aligned}$$

   Adding the three rows:

   $$y = x^5 - 5x^4 + 15x^3 - 26x^2 + 11x + 72.$$
   Writing one row per term of $u$ keeps the bookkeeping safe. Collect: $x^3$⁠: $7 + 8 = 15$⁠; $x^2$⁠: $9 - 35 = -26$⁠; $x$⁠: $-45 + 56 = 11$⁠.
   Distributive law. Basics: solving for dy/dx

2. Differentiate term by term

   $$\begin{aligned} \frac{dy}{dx} &= 5x^4 - 5\cdot 4x^3 + 15\cdot 3x^2 \\ &\quad - 26\cdot 2x + 11 + 0 \\ &= 5x^4 - 20x^3 + 45x^2 - 52x + 11. \end{aligned}$$
   Each term $ax^n$ becomes $nax^{n-1}$⁠; the constant $72$ disappears.
   Power rule (Table 5.3); sum rule (Section 5.3). Basics: derivative rules

Part (iii): by logarithmic differentiation

1. Where are the factors positive?

   $u = x^2 - 5x + 8$⁠: $a = 1 > 0$ and $b^2 - 4ac = 25 - 32 = -7 < 0$⁠, so $u > 0$ for every $x$⁠. Completing the square: $u = (x - \frac52)^2 + \frac74$⁠.

   $v = x^3 + 7x + 9$⁠: its derivative $3x^2 + 7$ is always positive, so $v$ is increasing and has at most one real zero. Since $v$ is continuous with $v(-2) = -13 < 0$ and $v(-1) = 1 > 0$⁠, it has a zero $r$ between $-2$ and $-1$ (intermediate value property); numerically $r = -1.0970811$⁠. So

   $$v > 0 \text{ for } x > r, \qquad v < 0 \text{ for } x < r.$$
   Logs exist only for positive numbers, so the log method applies directly where $y = uv > 0$⁠, that is for $x > r$⁠. Part (iii), Step 4 deals with $x \le r$⁠.
   Sign of a quadratic; positive derivative means increasing, so zero is crossed at most once (Theorem 1, Section 6.3, in the later Chapter 6); intermediate value property: a function continuous on $[a, b]$ takes every value between $f(a)$ and $f(b)$ (a standard property of continuous functions, not stated or proved in this chapter). Lesson: logarithmic differentiation

2. Take logs and differentiate (for $x > r$⁠)

   $$\log y = \log(x^2 - 5x + 8) + \log(x^3 + 7x + 9).$$

   Differentiate, using $\frac{d}{dx}\log g = \frac{g'}{g}$ on each term:

   $$\frac1y\frac{dy}{dx} = \frac{2x - 5}{x^2 - 5x + 8} + \frac{3x^2 + 7}{x^3 + 7x + 9}.$$
   The log of the product of two positive numbers is the sum of their logs. On the left, $\frac{d}{dx}\log y = \frac1y\frac{dy}{dx}$ by the chain rule.
   $\log pq = \log p + \log q$ (Section 5.4); Theorem 5 and the chain rule. Basics: exponentials and logarithms

3. Multiply by $y$ and simplify

   $$\begin{aligned} \frac{dy}{dx} = \ &(x^2 - 5x + 8)(x^3 + 7x + 9) \\ &\times\left[\frac{2x - 5}{x^2 - 5x + 8} + \frac{3x^2 + 7}{x^3 + 7x + 9}\right]. \end{aligned}$$

   Multiply each fraction by the product: each denominator cancels its own factor, leaving

   $$\begin{aligned} \frac{dy}{dx} &= (2x - 5)(x^3 + 7x + 9) \\ &\quad + (x^2 - 5x + 8)(3x^2 + 7), \end{aligned}$$

   which is exactly the expression of Part (i), Step 2. Expanding as in Part (i), Steps 3 and 4:

   $$\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11.$$
   The cancellation is allowed because both factors are non-zero for $x > r$⁠. Logarithmic differentiation of a product of two factors is the product rule in disguise.
   Logarithmic differentiation, $y' = y\cdot(\log y)'$ (Section 5.5). Lesson: logarithmic differentiation

4. The points $x \le r$

   For $x < r$⁠, $y < 0$⁠. Apply the method to $-y = u\cdot(-v) > 0$⁠: $\log(-y) = \log u + \log(-v)$⁠. Since $\frac{d}{dx}\log(-y) = \frac{-y'}{-y} = \frac{y'}{y}$ and $\frac{d}{dx}\log(-v) = \frac{-v'}{-v} = \frac{v'}{v}$⁠, we get the same equation as in Part (iii), Step 2, and so the same answer. At $x = r$ itself, $y = 0$ and no logarithm exists; there the answer comes from Part (i) or Part (ii), which hold for every $x$⁠, and it is the same polynomial.

   The book's warning that the function must be positive is about the logarithm, not about the derivative. Logging $-y$ (equivalently $|y|$⁠) extends the method wherever $y \ne 0$⁠.
   Chain rule on $\log(-g)$⁠: derivative $\frac{-g'}{-g} = \frac{g'}{g}$⁠. Basics: derivative rules

Do they all give the same answer?

1. Compare

   Part (i), Part (ii) and Part (iii) all give

   $$\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11.$$

   Yes: all three give the same answer, as they must, since a function has only one derivative. Logarithmic differentiation reaches it by a detour (logs exist only where $y \ne 0$⁠), but after simplification it is the product-rule expression, valid for every $x$⁠.

   For a product of only two polynomials, expanding (Part (ii)) is quickest; logs pay off for products of many factors or for powers with variable exponents.
   Uniqueness of the derivative: it is a single limit (Section 5.3). Basics: slope at a point

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