Question

For some constants a and b, find the derivative of
(i) (x-a)(x-b) (ii) (ax2+b)2 (iii)$\frac{x-a}{x-b}$

Answer 1

(i) Let f (x) = (x - a) (x + b)
f(x) = x2-(a+b)x+ab
∴ ƒ'(x) = $\frac{d}{dx}$(x2 - (a + b)x + ab)
=$\frac{d}{dx}$ (x2) -(a+b)$\frac{d}{dx}$(x) + $\frac{d}{dx}$(ab)
On using theorem $\frac{d}{dx}$(xn) = nxn-1, we obtain
f'(x)=2x-(a+b)+0=2x-a-b
(ii) Let f(x)=(ax2 +b)2
⇒f(x)=a2x4+2abx2+b2
∴ ƒ′(x) = (a2x4+2abx2+b2) = a2$\frac{d}{dx}$(x4)+2ab$\frac{d}{dx}$(x2)+$\frac{d}{dx}$(b2)
On using theorem $\frac{d}{dx}$ xn= nxn-1, we obtain
f'(x)=a2(4x3)+2ab (2x)+b2(0)
=4a2x3+4abx
=4ax(ax2+b)
(iii) Let f (x)=$\frac{(x-a)}{(x-b)}$
⇒f'(x)=$\frac{(x-a)}{(x-b)}$
By quotient rule,
$\frac{(x-b)\frac{d}{dx}(x-a)-(x-a)\frac{d}{dx}(x-b)}{(x-b)^2}$
= $\frac{(x-b)(1)-(x-a)(1)}{(x-b)^2}$
= $\frac{x-b-x+a}{(x-b)^2}$
=$\frac{a-b}{(x-b)^2}$