Question

Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed nonzero constants and m and n are integers): (ax +b)n (cx + d)m

Answer 1

Let f(x) = (ax +b)n (cx + d)m.
By Leibnitz product rule,
f'(x) = (ax+b)n$\frac{d}{dx}$ (cx + d)m + (cx +d)m $\frac{d}{dx}$(ax+b)n …...(1)
Now, let f1(x) = (cx+d)m
f1(x+h) = (cx+ch+d)m
f'(x) = $\lim_{h\rightarrow 0}$ $\frac{f(x+h)-f(x)}{h}$
= $\lim_{h\rightarrow 0}$ $\frac{(cx+ch+d)^m-(cx+dx)^m}{h}$
=(cx+dx)m $\lim_{h\rightarrow 0}$ $\frac{1}{h}$[(1+ $\frac{ch}{cx+1}$)m -1]
=(cx+dx)m$\lim_{h\rightarrow 0}$ $\frac{1}{h}$[($1+\frac{mch}{(cx+d)}$ + $\frac{m(m+1)}{2}$ $\frac{c^2h^2}{2(cx+d)^2}$ + ...) -1]
=(cx+dx)m$\lim_{h\rightarrow 0}$ $\frac{1}{h}$[$1+\frac{mch}{(cx+d)}$ + m(m-1)$\frac{c^2h^2}{2(cx+d)^2}$ + ....(Terms containing higher degrees of h)]
=(cx+dx)m $\lim_{h\rightarrow 0}$ $\frac{1}{h}$[$\frac{mc}{(cx+d)}$ + $\frac{m(m-1)c^2h^2}{2(cx+d)^2}$ + ...]
=(cx+dx)m [$\frac{mc}{(cx+d)}$+0]
= mc(cx+d)$\frac{m}{(cx+d)}$
= mc(cx+d)m-1 ....(2)
Similarly, $\frac{d}{dx}$(ax+b)n = na (ax+b) n-1 .....(3)
Therefore, From (1), (2), and (3), we obtain
f'(x) = (ax+b)n {mc (cx+d) m-1} + (cx+d)m {na(ax+b) n-1}
= (ax+b)n-1 (cx+d)m-1[mc(ax+b) + na(cx+d)]