Question

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

Answer 1

The correct answer is $∴\frac{dy}{dx}=\frac{du}{dx}.v.w+u.\frac{dv}{dx}.w+u.v.\frac{dw}{dx}$
Let $y=u.v.w=u.(v.w)$
By applying product rule, we obtain
$\frac{dy}{dx}=\frac{du}{dx}.(v.w)+u.\frac{d}{dx}(v.w)$
$⇒\frac{dy}{dx}=\frac{du}{dx}v.w+u[\frac{dv}{dx}.w+v.\frac{dw}{dx}]$ (Again applying product rule)
$⇒\frac{dy}{dx}=\frac{du}{dx}.v.w+u.\frac{dv}{dx}.w+u.v.\frac{dw}{dx}$
By taking logarithm on both sides of the equation $y=u.v.w$, we obtain
$logy=logu+logv+logw$
Differentiating both sides with respect to $x$, we obtain
$\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(logu)+\frac{d}{dx}(logv)+\frac{d}{dx}(logw)$
$⇒\frac{1}{y}.\frac{dy}{dx}=\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx}$
$⇒\frac{dy}{dx}=y(\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx})$
$⇒\frac{dy}{dx}=u.v.w(\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx})$
$∴\frac{dy}{dx}=\frac{du}{dx}.v.w+u.\frac{dv}{dx}.w+u.v.\frac{dw}{dx}$