For some constants a and b, find the derivative of

Solution:

(i) Let $f(x)=(x-a)(x-b)$

$\Rightarrow f(x)=x^{2}-(a+b) x+a b$

$\therefore f^{\prime}(x)=\frac{d}{d x}\left(x^{2}-(a+b) x+a b\right)$

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

On using theorem $\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$, we obtain

$f^{\prime}(x)=2 x-(a+b)+0=2 x-a-b$

(ii) Let $f(x)=\left(a x^{2}+b\right)^{2}$

$\Rightarrow f(x)=a^{2} x^{4}+2 a b x^{2}+b^{2}$

$\therefore f^{\prime}(x)=\frac{d}{d x}\left(a^{2} x^{4}+2 a b x^{2}+b^{2}\right)=a^{2} \frac{d}{d x}\left(x^{4}\right)+2 a b \frac{d}{d x}\left(x^{2}\right)+\frac{d}{d x}\left(b^{2}\right)$

On using theorem $\frac{d}{d x} x^{n}=n x^{n-1}$, we obtain

$f^{\prime}(x)=a^{2}\left(4 x^{3}\right)+2 a b(2 x)+b^{2}(0)$

$=4 a^{2} x^{3}+4 a b x$

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

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

$\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(\frac{x-a}{x-b}\right)$

By quotient rule,

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

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

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

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