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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.
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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