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If $u, v$ and $w$ are functions of $w$ then show that $\frac{d}{d x}$ (u. v. w) $=\frac{d u}{d x} \mathrm{v} \cdot \mathrm{w}+\mathrm{u} \cdot \frac{d v}{d x} \cdot \mathrm{w}+\mathrm{u} . \mathrm{v} \frac{d w}{d x}$ in two waysfirst by repeated application of product rule, second by logarithmic differentiation.
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Let $y=u . v . w \Rightarrow y=u \cdot(v w)$
Differentiating w.r.t. $\mathrm{x}$
$\begin{aligned}
&\text { (i) } \frac{d y}{d x}=\frac{d u}{d x} \cdot(v w)+u \frac{d}{d x}(v w) \\
&=u^{\prime} \cdot(v w)+u\left[v^{\prime} w+v^{\prime}\right] \\
&=u^{\prime} \cdot v \cdot w+u v^{\prime} w+u v w^{\prime} \\
&\quad \therefore \frac{d y}{d x}=\frac{d u}{d x} v \cdot w+u \cdot \frac{d v}{d x} \cdot w+u \cdot v \cdot \frac{d w}{d x}
\end{aligned}$
(ii) $\mathrm{y}=\mathrm{u}$. V. w; Taking log of both sides, $\log y=\log u+\log v+\log w$
Differentiating w.r.t. $x$
$\begin{aligned}
&\frac{1}{y} \frac{d y}{d x}=\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x} \\
&\therefore \frac{d y}{d x}=\frac{d u}{d x} . v . w+u \cdot \frac{d v}{d x} \cdot w+u . v . \frac{d w}{d x}
\end{aligned}$
Differentiating w.r.t. $\mathrm{x}$
$\begin{aligned}
&\text { (i) } \frac{d y}{d x}=\frac{d u}{d x} \cdot(v w)+u \frac{d}{d x}(v w) \\
&=u^{\prime} \cdot(v w)+u\left[v^{\prime} w+v^{\prime}\right] \\
&=u^{\prime} \cdot v \cdot w+u v^{\prime} w+u v w^{\prime} \\
&\quad \therefore \frac{d y}{d x}=\frac{d u}{d x} v \cdot w+u \cdot \frac{d v}{d x} \cdot w+u \cdot v \cdot \frac{d w}{d x}
\end{aligned}$
(ii) $\mathrm{y}=\mathrm{u}$. V. w; Taking log of both sides, $\log y=\log u+\log v+\log w$
Differentiating w.r.t. $x$
$\begin{aligned}
&\frac{1}{y} \frac{d y}{d x}=\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x} \\
&\therefore \frac{d y}{d x}=\frac{d u}{d x} . v . w+u \cdot \frac{d v}{d x} \cdot w+u . v . \frac{d w}{d x}
\end{aligned}$
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