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

(i) Let $y = u \cdot v \cdot w = u \cdot (vw)$ ...(i)
Differentiating (i) on both sides w.r.t. x, we get

$\cfrac{{dy}}{{dx}} = \cfrac{d}{{dx}}u \cdot (vw) = u\cfrac{d}{{dx}}(vw) = u' \cdot (vw) + u[v'w + vw']$

$= u' \cdot v \cdot w + uv'w = \cfrac{{du}}{{dx}} \cdot v \cdot w + u \cdot \cfrac{{dv}}{{dx}} \cdot w + u \cdot v \cdot \cfrac{{dw}}{{dx}}$

(ii) $y = u \cdot v \cdot w$
By taking log on both sides , we get
$\log y = \log u + \log v + \log w$ ….(ii)
Differentiating (ii) on both sides w.r.t. x, we get

$\cfrac{1}{y}\cfrac{{dy}}{{dx}} = \cfrac{1}{u}\cfrac{{du}}{{dx}} + \cfrac{1}{v}\cfrac{{dv}}{{dx}} + \cfrac{1}{w}\cfrac{{dw}}{{dx}}$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = y\left( {\cfrac{1}{u}\cfrac{{du}}{{dx}} + \cfrac{1}{v}\cfrac{{dv}}{{dx}} + \cfrac{1}{w}\cfrac{{dw}}{{dx}}} \right)$

$= uvw\left( {\cfrac{1}{u}\cfrac{{du}}{{dx}} + \cfrac{1}{v}\cfrac{{dv}}{{dx}} + \cfrac{1}{w}\cfrac{{dw}}{{dx}}} \right)$

$= vw\cfrac{{du}}{{dx}} + uw\cfrac{{dv}}{{dx}} + uv\cfrac{{dw}}{{dx}} = \cfrac{{du}}{{dx}} \cdot v \cdot w + u \cdot \cfrac{{dv}}{{dx}} \cdot w + u \cdot v \cdot \cfrac{{dw}}{{dx}}$ .