Maths

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Kindly go through the solution

56. Given, f (x) = (ax + b)ⁿ (cx + d)^m

So, f'(x) = ${(ax+b)}^n\frac{d}{dx}{(cx+d)}^n+{(cx+d)}^m\frac{d}{dx}{(ax+b)}^n$

$={(ax+b)}^x·mc{(cx+d)}^{m-1}+{(cx+d)}^{m}na{(ax+b)}^{x-1}$

$={(ax+b)}^{n-1}(cx+d){}^{m-1}[(ax+b)·mc+(cx+d)na]$

$={(ax+b)}^{n-1}(cx+d){}^{m-1}[mc(ax+b)+na(cx+d)]$

55. Given, f (x) = (ax + b)ⁿ

Chain rule,  $\frac{d}{dx}u{(x)}^n=nu{(x)}^{n-1}\frac{du}{dx}u(x)$ where

u (x) is a function of x.

So, f (x) = $\frac{d}{dx}{(ax+b)}^n$

$=n{(ax+b)}^{n-1}\frac{d}{dx}(ax+b)$

$=na{(ax+b)}^{n-1}$

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution