1 Answer
Solution :-
Let f(x) = (ax + b)n. Accordingly,
f(x + h) = {a(x + h) + b}n = (ax + ah + b)n By first principle,
\( f'(x)=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h} \)
\( \lim\limits_{h \to 0}\dfrac{(ax+ah+b)-(ax+b)^n}{h} \)
_1640246994.png)
_1640247118.png)
(using binomial theorem)
_1640247269.png)
(Terms containing higher degrees of h)
_1640247388.png)
\( (ax+b)^n[\dfrac{na}{(ax+b)}+0]\)
\( na\dfrac{(ax+b)^n}{ax+b}\)
\(na(ax+b)^{n-1}\)
Answered by Abhisek | 1 year ago