Exercise 5.5 · Q18: Derivative of $uvw$ two ways

Basics you need

Part (i): by repeated use of the product rule

1. Treat $uvw$ as a product of two factors

   Write $y = (uv)\cdot w$⁠. The first factor is $uv$⁠, the second is $w$⁠. By the product rule,

   $$\frac{dy}{dx} = \frac{d}{dx}(uv)\cdot w + (uv)\cdot\frac{dw}{dx}.$$
   The product rule is stated for two factors. Grouping $u$ and $v$ together makes the three-factor product a two-factor one. The grouping is allowed because multiplication is associative: $(uv)w = uvw$⁠.
   Product (Leibnitz) rule (Section 5.3). Basics: derivative rules

2. Use the product rule again, on $uv$

   $$\frac{d}{dx}(uv) = \frac{du}{dx}\,v + u\,\frac{dv}{dx}.$$
   $uv$ is itself a product of two differentiable functions, so the same rule applies. It is differentiable because $u$ and $v$ are.
   Product rule (Section 5.3). Basics: derivative rules

3. Substitute and expand

   $$\begin{aligned} \frac{dy}{dx} &= \left(\frac{du}{dx}\,v + u\,\frac{dv}{dx}\right)w + uv\,\frac{dw}{dx} \\ &= \frac{du}{dx}\,vw + u\,\frac{dv}{dx}\,w + uv\,\frac{dw}{dx}. \end{aligned}$$

   This is the required formula, and it holds at every $x$ where $u$⁠, $v$⁠, $w$ are differentiable.

   Multiplying the bracket by $w$ (distributive law) gives the first two terms; the third comes from the second half of Part (i), Step 1. Nothing here needs $u$⁠, $v$ or $w$ to be non-zero.
   Distributive law. Basics: solving for dy/dx

Part (ii): by logarithmic differentiation

1. Take logs (where $u, v, w > 0$⁠)

   At points where $u$⁠, $v$⁠, $w$ are all positive, $y = uvw > 0$ and

   $$\log y = \log u + \log v + \log w.$$
   Logarithmic differentiation needs the quantities whose logs we take to be positive (Section 5.5). The log of a product of positive numbers is the sum of their logs.
   $\log pq = \log p + \log q$⁠, applied twice (Section 5.4). Basics: exponentials and logarithms

2. Differentiate both sides

   Each term is the log of a function of $x$⁠, so by the chain rule

   $$\frac1y\frac{dy}{dx} = \frac1u\frac{du}{dx} + \frac1v\frac{dv}{dx} + \frac1w\frac{dw}{dx}.$$
   Outer function $\log$⁠, inner function $u$ (or $v$⁠, $w$⁠, $y$⁠): $\frac{d}{dx}\log u = \frac1u\cdot\frac{du}{dx}$⁠.
   Theorem 5 and the chain rule (Theorem 4, Section 5.3.1). Lesson: logarithmic differentiation

3. Multiply by $y = uvw$

   $$\begin{aligned} \frac{dy}{dx} &= uvw\left[\frac{1}{u}\frac{du}{dx} + \frac{1}{v}\frac{dv}{dx} + \frac{1}{w}\frac{dw}{dx}\right] \\ &= \frac{du}{dx}\,vw + u\,\frac{dv}{dx}\,w + uv\,\frac{dw}{dx}. \end{aligned}$$
   In the first term $uvw\cdot\frac1u = vw$⁠; in the second $uvw\cdot\frac1v = uw$⁠; in the third $uvw\cdot\frac1w = uv$⁠. This is the same formula as Part (i), Step 3.
   Logarithmic differentiation, $y' = y\cdot(\log y)'$ (Section 5.5). Lesson: logarithmic differentiation

4. When some of $u, v, w$ are negative, or zero

   Negative but non-zero. Replace each negative factor $g$ by $-g$⁠, which is positive. Then $|y| = |u|\,|v|\,|w|$ and

   $$\log|y| = \log|u| + \log|v| + \log|w|.$$

   Since $\frac{d}{dx}\log(-g) = \frac{-g'}{-g} = \frac{g'}{g}$⁠, each term still differentiates to $\frac{g'}{g}$⁠, so Part (ii), Steps 2 and 3 go through unchanged.

   Zero. At a point where one of $u$⁠, $v$⁠, $w$ is $0$⁠, $y = 0$ and no logarithm exists, so the log method says nothing there. The formula is still true at such points by Part (i), which never divided by $u$⁠, $v$ or $w$⁠.

   This is why Part (i) is the real proof for all points, while Part (ii) is a quick derivation wherever the three functions are non-zero. Both give the same formula.
   Chain rule; positivity requirement (Section 5.5). Basics: derivative rules

Every fact used, and where it comes from