Question 18: 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.
Class 12 - Math - Continuity and Differentiability
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we have to prove
d/dx(u.v.w)= du/dx v.w + u. dv/dx .w + u. v dw/dx
By product rule
let u.v = k
now d/dx(k.w) = kd/dx(w)+wd/dx(k)
also d/dx(w) = d/dx(v.u) = vd/dx(u)+ud/dx(v)
=>d/dx(u.v.w) =du/dx v.w + u. dv/dx .w + u. v dw/dx
Now using logarithmic differentiation
let y = v.u.w
log y =logv + log u + log w
differentiating w.r.t x
1/y dy/dx = 1/v dv/dx + 1/udu/dx + 1/w . dw/dx
=> d(u.v.w)/dx =du/dx v.w + u. dv/dx .w + u. v dw/dx
Hence proved
d/dx(u.v.w)= du/dx v.w + u. dv/dx .w + u. v dw/dx
By product rule
let u.v = k
now d/dx(k.w) = kd/dx(w)+wd/dx(k)
also d/dx(w) = d/dx(v.u) = vd/dx(u)+ud/dx(v)
=>d/dx(u.v.w) =du/dx v.w + u. dv/dx .w + u. v dw/dx
Now using logarithmic differentiation
let y = v.u.w
log y =logv + log u + log w
differentiating w.r.t x
1/y dy/dx = 1/v dv/dx + 1/udu/dx + 1/w . dw/dx
=> d(u.v.w)/dx =du/dx v.w + u. dv/dx .w + u. v dw/dx
Hence proved
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