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)]