Solution:

Provided $u, v$ and $w$ are the functions of $x$.

We need to prove: $\frac{d}{d x}(u \cdot v \cdot w)=\frac{d u}{d x} \cdot v \cdot w+u \cdot \frac{d v}{d x} \cdot w+u \cdot v \cdot \frac{d w}{d x}$

First way: By repeated application of product rule

Left Hand Side (L.H.S.)

$\frac{d}{d x}(u . v \cdot w)=\frac{d}{d x}[(w v) \cdot w]$

$=u v \frac{d}{d x} w+w \frac{d}{d x}(u v)$

$=u v \frac{d w}{d x}+w\left[u \frac{d}{d x} v+v \frac{d}{d x} u\right.$

$=u v \frac{d w}{d x}+u w \frac{d v}{d x}+v w \frac{d u}{d x}$

$=\frac{d u}{d x} \cdot v \cdot w+u \cdot \frac{d v}{d x} \cdot w+u \cdot v \cdot \frac{d w}{d x}$

= Right Hand Side (R.H.S.)

As a result, proved.

Second way: By Logarithmic differentiation

Let’s take $y=u v$

$\log y=\log (u . v . w)$

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

$\frac{d}{d x} \log y=\frac{d}{d x} \log u+\frac{d}{d x} \log v+\frac{d}{d x} \log w$

$\frac{1}{y} \frac{d y}{d x}=\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x}$

$\frac{d y}{d x}=y\left[\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x}\right]$

By putting the value of y = uvw, we obtain

$\frac{d}{d x}(u . v . w)=u w\left[\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x}\right]$

$\frac{d}{d x}(u . v \cdot w)=\frac{d u}{d x} \cdot v w+u \cdot \frac{d v}{d x} \cdot w+u . v \cdot \frac{d w}{d x}$

As a result, proved.