Q.If $u, v$ and $w$ are functions of $x$, then show that $\frac{d}{dx} (u \cdot v \cdot w) = \frac{du}{dx} \cdot v \cdot w + u \cdot \frac{dv}{dx} \cdot w + u \cdot v \cdot \frac{dw}{dx}$ in two ways - first by repeated application of product rule, second by logarithmic differentiation.

Concept: Chain Rule & Product Rule

Method 1 — Repeated Product Rule

Treat $u \cdot v \cdot w$ as $(u \cdot v) \cdot w$.

First apply product rule to $u \cdot v$:

$\frac{d}{dx}(u v) = u' v + u v'$.

Now differentiate the product with $w$:

$\frac{d}{dx}[(u v) w] = (u' v + u v') w + (u v) w'$

$= u' v w + u v' w + u v w'$.

Method 2 — Logarithmic Differentiation

Let $y = u v w$. Take $\log$ on both sides:

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

Differentiate with respect to $x$:

$\frac{1}{y} \frac{dy}{dx} = \frac{u'}{u} + \frac{v'}{v} + \frac{w'}{w}$. …

Why this works — the Chain Rule in disguise

The product rule for two functions says:

$\frac{d}{dx}(u v) = u' v + u v'$.

For three functions, we are really just applying that rule twice. Think of $u \cdot v \cdot w$ as $(u \cdot v) \cdot w$ — a product of two "functions", where the first factor is itself a product. The Chain Rule never appears directly here (no composition), but the idea of "differentiate one factor at a time" is the same pattern that extends to any number of factors.

Logarithmic differentiation gives an alternative view: it turns multiplication into addition, so the derivative becomes a sum of individual logarithmic derivatives — which is exactly the same result.

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Method 1: Repeated application of the product rule

Step 1. Treat $u \cdot v \cdot w$ as $(u v) \cdot w$.

Let $p = u v$. Then we have $\frac{d}{dx}(p \cdot w)$.

Step 2. Apply the product rule to $p \cdot w$:

Step 3. Now $\frac{dp}{dx} = \frac{d}{dx}(u v)$. Apply the product rule again:

Step 4. Substitute back:

Step 5. Expand:

That's the result.

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Method 2: Logarithmic differentiation

Step 1. Assume $u, v, w > 0$ (so logs are defined). Take the natural log of both sides:

Step 2. Differentiate both sides with respect to $x$. On the left, by the Chain Rule:

On the right, differentiate term by term: …

Method: Extending a Two-Factor Rule to Three Factors

When you're asked to derive a formula for three (or more) functions multiplied together, the general strategy is to reduce it to a rule you already know, applied more than once — never derive it from scratch with limits.

Steps

Step 1 (Method A — repeated product rule): Group two factors together first

Treat $u\cdot v\cdot w$ as $(u\cdot v)\cdot w$ — a product of two things, where the first "thing" happens to itself be a product. This lets you apply the ordinary two-factor product rule you already know, twice.

Step 2: Apply the product rule to the outer grouping

Step 3: Apply the product rule again to expand $\dfrac{d}{dx}(uv)$, then substitute back

Substituting gives the three-term result — each term differentiates exactly one of $u,v,w$ while leaving the other two unchanged.

Step 4 (Method B — logarithmic differentiation): an independent cross-check …

Common Mistakes

Mistake 1: Writing only two terms instead of three when generalising the product rule

Why it's wrong: a very common error when extending the two-factor product rule to three factors is to write $\frac{d}{dx}(uvw) = u'vw + uvw'$ (skipping the middle term $uv'w$) — an easy slip because the two-factor rule only has two terms to pattern-match against. Correct approach: apply the two-factor rule twice, in stages (first to $(uv)\cdot w$, then to $u\cdot v$ inside), so no term gets silently dropped.

Mistake 2: Using logarithmic differentiation without noting the positivity requirement …