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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.

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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}=$$y\bigg(\large\frac{1}{u}.\frac{du}{dx}+\frac{1}{v}.\frac{dv}{dx}+\frac{1}{w}.\frac{dw}{dx}\bigg)$
Substitute the value of $y$
$\large\frac{dy}{dx}=$$uvw\bigg(\large\frac{1}{u}.\frac{du}{dx}+\frac{1}{v}.\frac{dv}{dx}+\frac{1}{w}.\frac{dw}{dx}\bigg)$
$\quad\;=uvw.\large\frac{1}{u}.\frac{du}{dx}+$$uvw.\large\frac{1}{v}.\frac{dv}{dx}+$$uvw.\large\frac{1}{w}.\frac{dw}{dx}$
$\quad\;=vw.\large\frac{du}{dx}+$$uw.\large\frac{dv}{dx}+$$uv.\large\frac{dw}{dx}$
answered May 9, 2013 by sreemathi.v
edited May 9, 2013 by sreemathi.v
 

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