Answer
a) $\frac{-2}{\sqrt{9-4x}}$
b) $\frac{1}{(x+1)^2}$
Work Step by Step
a) We have, $\frac{d\sqrt{9-4x}}{dx}$
$\quad = \frac{1}{2\sqrt{9-4x}}\times \frac {d(9-4x)}{dx}\quad$ (Chain rule)
$\quad = \frac{-2}{\sqrt{9-4x}}\quad\quad (\because\frac {d(9-4x)}{dx}=-4)$
b) We have, $\frac{d(\frac{x}{x+1})}{dx}$
Let $x = U, x+1=V$
By quotient rule,
$\implies \frac{dy}{dx} = \frac{U'V-V'U}{V^2} = \frac{(1)(x+1)-(1)(x)}{(x+1)^2} = \frac{1}{(x+1)^2}$
Checking result using derivative formula:
$\frac {d(f(x))}{dx} = \lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$
a) $\frac {d(f(x))}{dx} = \lim\limits_{h \to 0} \frac{\sqrt{9-4(x+h)}-\sqrt{9-4x}}{h}$
Rationalizing numerator,
$\frac {d(f(x))}{dx} = \lim\limits_{h \to 0} \frac{\sqrt{9-4(x+h)}-\sqrt{9-4x}}{h}\times\frac{\sqrt{9-4(x+h)}+\sqrt{9-4x}}{{\sqrt{9-4(x+h)}+\sqrt{9-4x}}} = \lim\limits_{h \to 0}\frac{-4h}{h({\sqrt{9-4(x+h)}+\sqrt{9-4x})}}$
Cancelling $h$ and applying limit,
$\implies \frac{-4}{({\sqrt{9-4(x)}+\sqrt{9-4x})}} = \frac{-4}{2{\sqrt{9-4(x)}}}=\frac{-2}{\sqrt{9-4x}}$
$\therefore Verified.$
b) $\frac {d(f(x))}{dx} = \lim\limits_{h \to 0} \frac{\frac{x+h}{x+h+1}-\frac{x}{x+1}}{h}$
Taking LCM of numerator,
$\frac{d(f(x))}{dx} = \frac{\frac{(x+h)(x+1)-x(x+h+1)}{(x+h+1)(x+1)}}{h} = \frac{(x+h)(x+1)-x(x+h+1)}{(h)(x+h+1)(x+1)} = \frac{h}{(h)(x+h+1)(x+1)}$
Taking limit and cancelling h,
$\implies \frac{1}{(x+1)(x+1)} = \frac{1}{(x+1)^2}$
$\therefore Verified.$
