If u - v and w are functions of x- then show that d - dx u - v - w - du - dx v - w - u - dv - dx w - u - v dw - dx In two ways-first by repeated application of product rule-second by logarithmic differentiation-

First, we use product rule and find that d/dx(uvw)=d/dx{(uv)w}=(uv)dw/dx+wd/dx(uv) (Consider here uv as one function and w as another function) =uvdw/dx+w(udv ...

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Byju's Answer

Standard XII

Mathematics

Instantaneous Rate of Change

If u , v and ...

Question

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

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Solution

First, we use product rule and find that

$\frac{d}{d x} (u v w) = \frac{d}{d x} {(u v) w} = (u v) \frac{d w}{d x} + w \frac{d}{d x} (u v) (C o n s i d e r h e r e u v a s o n e f u n c t i o n a n d w a s a n o t h e r f u n c t i o n) = u v \frac{d w}{d x} + w (u \frac{d v}{d x} + v \frac{d u}{d x}) = u v \frac{d w}{d x} + w u \frac{d v}{d x} + w v \frac{d u}{d x} . . . . . (i) N e x t, w e u s e l o g a r i t h m i c d i f f e r e n t i a t i o n t o o b t a i n d / d x (u v w) . L e t y = u v w \Rightarrow l o g y = l o g (u v w) l o g y = l o g u + l o g v + l o g w D i f f e r e n t i a t i n g w . r . t, w e g e t \frac{1}{y} \frac{d y}{d x} = \frac{1}{u} \frac{d u}{d x} + \frac{1}{v} \frac{d v}{d x} + \frac{1}{w} \frac{d w}{d x} \Rightarrow \frac{d y}{d x} = y {\frac{1}{u} \frac{d u}{d x} + \frac{1}{v} \frac{d v}{d x} + \frac{1}{w} \frac{d w}{d x}} = (u v w) {\frac{1}{u} \frac{d u}{d x} + \frac{1}{v} \frac{d v}{d x} + \frac{1}{w} \frac{d w}{d x}} = (v w) \frac{d u}{d x} + (u w) \frac{d v}{d x} + (u v) \frac{d w}{d x} = u v \frac{d w}{d x} + u w \frac{d v}{d x} + w v \frac{d u}{d x} . . . . . . (i i)$

From Eqs. (i) and (ii), we find that the result obtained is same in the two cases.

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