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

Let $y = u \cdot v \cdot w = u \cdot (v \cdot w).$
Method $1$: By applying the product rule repeatedly:
$\frac{dy}{dx} = \frac{du}{dx} \cdot (v \cdot w) + u \cdot \frac{d}{dx}(v \cdot w)$
$= \frac{du}{dx} \cdot v \cdot w + u \cdot \left[ \frac{dv}{dx} \cdot w + v \cdot \frac{dw}{dx} \right]$
$= \frac{du}{dx} \cdot v \cdot w + u \cdot \frac{dv}{dx} \cdot w + u \cdot v \cdot \frac{dw}{dx}.$
Method $2$: By logarithmic differentiation:
Taking the natural logarithm on both sides of $y = u \cdot v \cdot w,$ we get
$\log y = \log u + \log v + \log w.$
Differentiating both sides with respect to $x,$ we obtain
$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} + \frac{1}{v} \cdot \frac{dv}{dx} + \frac{1}{w} \cdot \frac{dw}{dx}.$
Multiplying both sides by $y,$ we get
$\frac{dy}{dx} = y \left( \frac{1}{u} \cdot \frac{du}{dx} + \frac{1}{v} \cdot \frac{dv}{dx} + \frac{1}{w} \cdot \frac{dw}{dx} \right).$
Substituting $y = u \cdot v \cdot w,$ we obtain
$\frac{dy}{dx} = (u \cdot v \cdot w) \left( \frac{1}{u} \cdot \frac{du}{dx} + \frac{1}{v} \cdot \frac{dv}{dx} + \frac{1}{w} \cdot \frac{dw}{dx} \right)$
$= \frac{du}{dx} \cdot v \cdot w + u \cdot \frac{dv}{dx} \cdot w + u \cdot v \cdot \frac{dw}{dx}.$