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.

If u, v and w are functions of x, then show that`d/(dx)(udotvdotw)=(d

1.	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.
Answer» `(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.