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# If $u,\: v$ and $w$ are functions of $x$, then show that $\large\frac{d}{dx} $$( u.v.w) =\large \frac{du}{dx}$$v.w + u . \large\frac{dv}{dx} $$. w + u.v. \large\frac{dw}{dx} in two ways - first by repeated application of product rule, second by logarithmic differentiation. Can you answer this question? ## 1 Answer 0 votes Toolbox: • Product rule y=u.v.w=u.(vw) • \log mn=\log m+\log n Step 1: y=u.v.w Differentiating on both sides (i) \large\frac{dy}{dx}$$=u'(vw)+u\large\frac{d}{dx}$$(vw) \qquad=u'(vw)+u[v'w+vw'] \qquad\;=u'v.w+uv'w+uvw' \qquad\;=\large\frac{du}{dx}$$.v.w+u.\large\frac{dv}{dx}$$.w+u.v.\large\frac{dw}{dx} y=u.v.w Step 2: (ii) Taking \log on both sides \log y=\log u+\log v+\log w Differentiating on both sides \large\frac{1}{y}.\frac{dy}{dx}=\frac{1}{u}.\frac{du}{dx}+\frac{1}{v}.\frac{dv}{dx}+\frac{1}{w}.\frac{dw}{dx} \large\frac{dy}{dx}=$$y\bigg(\large\frac{1}{u}.\frac{du}{dx}+\frac{1}{v}.\frac{dv}{dx}+\frac{1}{w}.\frac{dw}{dx}\bigg)$
Substitute the value of $y$