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

Answer 1

Let f(x) = (ax +b)ⁿ. Accordingly, f(x+h) = {a(x+h) + b}ⁿ = (ax + ah +b )ⁿ
by first principle,
f'(x) = \(\lim_{h\rightarrow 0}\) \(\frac{f(x+h)-f(x)}{h}\)
=\(\lim_{h\rightarrow 0}\) (ax+ah+b)ⁿ -(ax+b)ⁿ
=\(\lim_{h\rightarrow 0}\) \(\frac{(ax+b)^n(1+\frac{ah}{ax+b})^n-(ax+b)^n}{h}\)
=(ax+b)ⁿ \(\lim_{h\rightarrow 0}\)\(\frac{1}{n}\)[{ 1+n(\(\frac{ah}{ax+b}\)) +\(\frac{n(n-1)}{2}\)(\(\frac{ah}{ax+b}\))² + .....}-1](Using binomial theorem)
=(ax+b)ⁿ \(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[n(\(\frac{ah}{ax+b}\)) + n(n-1)\(\frac{a^2h^2}{(ax+b)^2}\) + .....(Terms containing higher degrees of h)]
=(ax+b)ⁿ\(\lim_{h\rightarrow 0}\)  \(\frac{1}{h}\)[ \(\frac{na}{ax+b}\)+ n(n-1)\(\frac{a^2h^2}{(ax+b)^2}\) + ..]
=(ax+b)ⁿ [\(\frac{na}{(ax+b)}\) + 0]
=na\(\frac{(ax+b)^n}{(ax+b)}\)
=na (ax+b)^n-1