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) = nx^n-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= nx^n-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,
\(\frac{(x-b)\frac{d}{dx}(x-a)-(x-a)\frac{d}{dx}(x-b)}{(x-b)^2}\)
= \(\frac{(x-b)(1)-(x-a)(1)}{(x-b)^2}\)
= \(\frac{x-b-x+a}{(x-b)^2}\)
=\(\frac{a-b}{(x-b)^2}\)