If u, v and w are functions of x, then show that

Let $y=u \cdot v \cdot w=u \cdot(v \cdot w)$

By applying product rule, we obtain

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

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x} v \cdot w+u\left[\frac{d v}{d x} \cdot w+v \cdot \frac{d w}{d x}\right] \quad$ (Again applying product rule)

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

By taking logarithm on both sides of the equation $y=u, v \cdot w$, we obtain

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

Differentiating both sides with respect to x, we obtain

$\frac{1}{y} \cdot \frac{d y}{d x}=\frac{d}{d x}(\log u)+\frac{d}{d x}(\log v)+\frac{d}{d x}(\log w)$

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

$\Rightarrow \frac{d y}{d x}=y\left(\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x}\right)$

$\Rightarrow \frac{d y}{d x}=u . v . w \cdot\left(\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x}\right)$

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