show directly(without using the chain Rule) that [f(cx)]´=cf´(cx) for all c in R?
Let f:R-->R be differentiable.Show directly(without using the chain Rule) that [f(cx)]´=cf´(cx) for all c in R
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d/dx(f(cx))
= lim(h --> 0) [f(c(x + h)) - f(cx)]/h
= lim(h ---> 0) [f(cx + ch) - f(cx)]/h
= lim(u ---> 0) [f(cx + u) - f(cx)]/(u/c) (u = h/c -->0 as h --> 0)
= c lim(u ---> 0) [f(cx + u) - f(cx)]/u
= c f'(cx).