Question

For some constants a and b, find the derivative of:

(i) (x-a)(x-b)

(ii) $(a{x}^2+b{)}^2$

(iii) $\frac{x-a}{x-b}$

Answer 1

(i) Here f(x) = (x-a)(x-b)

$\therefore$ f'(x)= $\frac{d}{dx}$ (x-a)(x-b)

= $(x-a)\frac{d}{dx}(x-b)+(x-b)\frac{d}{dx}(x-a)$

= $(x-a)\times 1+(x-b)\times 1$

= x-a+x-b= 2x -a-b.

(ii) Here f(x) = $(a{x}^2+b{)}^2=a^2{x}^4+b^2+2ab{x}^2$

$\therefore$ f'(x) = $\frac{d}{dx}[a^2{x}^4+b^2+2ab{x}^2]$

= $a^2\frac{d}{dx}\left(x^4\right)+\frac{d}{dx}\left(b^2\right)+2ab\frac{d}{dx}\left(x^2\right)$

= $a^2\times 4x^3+0+2ab\times 2x$

= $4a^2{x}^3+4abx$.

= 4ax$(a{x}^2+b)$.

(iii) Here f(x)= $\frac{x-a}{x-b}$

$\therefore$ f'(x) = $\frac{d}{dx}\left(\frac{x-a}{x-b}\right)$

= $\frac{(x-b)\frac{d}{dx}(x-a)-(x-a)\frac{d}{dx}(x-b)}{(x-b{)}^2}$

= $\frac{(x-b)\times 1-(x-a)\times 1}{(x-b{)}^2}=\frac{x-b-x+a}{(x-b{)}^2}$

= $\frac{a-b}{(x-b{)}^2}$.