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प्रश्न
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
उत्तर
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)]
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