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If u, v and w are functions of x, then s...

If u, v and w are functions of x, then show that `d/(dx)(udotvdotw)=(d u)/(dx)vdotw+udot(d v)/(dx)dotw+udotv(d w)/(dx)` in two ways - first by repeated application of product rule, second by logarithmic differentiation.

Text Solution

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`(i) (dy)/(dx) = (d)/(dx) {(uv)w}`
`= w * (d)/(dx) (uv) + uv * (d)/(dx)(w)`
`= w[v(du)/(dx) + u(dv)/(dx)] + uv(dw)/(dx)`
`=vw (du)/(dx) + uw (dv)/(dx) + uv(dw)/(dx)`
(ii) y = uvw
`rArr "log"y = "log (uvw)`
`="log" u + "log"v + "log"w`
Differentiate both sides w.r.t.x
`(1)/(y) (dy)/(dx) = (1)/(u) (du)/(dx) + (1)/(v) (dv)/(dx) + (1)/(w) (dw)/(dx)`
`rArr (dy)/(dx) = (y)/(u) (du)/(dx) + (y)/(v) (dv)/(dx) + (y)/(w) (dw)/(dx)`
`= (uvw)/(u) (du)/(dx) + (uvw)/(v) (dv)/(dx) + (uvw)/(w) (dw)/(dx)`
`= vw(du)/(dx) + uw(dv)/(dx) + uv(dw)/(dx)`
Thus, the results in two cases are same.
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