Question

Find the derivative of the following function (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero 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 d/dx (cx + d)^m + (cx + d)^m d/dx (ax + b)^n`    ...(1)

Now, let f₁(x) = (cx + d)^m

f₁(x + h) = (cx + ch + d)^m

f₁(x) = `lim_(h->0)(f_1(x + h) - f_1(x))/h`

= `lim_(h->0) ((cx + ch + d)^m - (cx + d)^n)/h`

= `(cx + d)^m lim_(h-0)1/h [(1 + (ch)/(cx + d))^m - 1]`

= `(cx + d)^m lim_(h-0) 1/h[(1 + (mch)/(cx + d) + (m(m - 1))/2 ((c^2h^2))/(cx + d)^2 + ...) -1]`

= `(cx + d)^m lim_(h->0) 1/h [(mch)/(cx + d) + (m(m - 1)c^2h^2)/(2(cx + d)^2) + ...("Terms containing higher degrees of h")]`

= `(cx + d)^m lim_(h->0) [(mc)/(cx + d) + (m(m - 1)c^2h)/(2(cx + d)^2 + ...]]`

= `(cx + d)^m [(mc)/(cx + d) + 0]`

= `(mc(cx + d)^m)/(cx + d)`

= mc (cx + d)^{m - 1}

`d/dx (cx + d)^m` = mc (cx + d)^{m - 1}      .....(2)

Similarly, `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)]