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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 - Mathematics

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Question

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.

Sum
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Solution

Let y = u. v. w = u. (vw)   ....(i)

Differentiating (i) both sides w.r.t. x, we get

(i) `dy/dx = u' .(vw) + u d/dx (vw)`

= u'. (vw) + u [v' w + vw']

= u'. v. w + uv w + uvw'

= `(du)/dx. v. w + u. (dv)/dx . w + u.v. (dw)/dx`

(ii) y = u. v. w

Taking log on both sides, we get

log y = log u + log v + log w   ....(ii)

Differentiating (ii) both sides w.r.t. x, we get

`1/y dy/dx = 1/u (du)/dx + 1/v (dv)/dx + 1/w (dw)/dx`

`dy/dx = y (1/u (du)/dx + 1/v (dv)/dx + 1/w (dw)/dx)`

= `uvw (1/u (du)/dx + 1/v (dv)/dx + 1/w (dw)/dx)`

= `vw (du)/dx + uw (dv)/dx + uv (dw)/dx`

= `(du)/dx. v. w + u. (dv)/dx .w + u. v (dw)/dx`

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Chapter 5: Continuity and Differentiability - Exercise 5.5 [Page 179]

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NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 5 Continuity and Differentiability
Exercise 5.5 | Q 18 | Page 179

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