[tex]\textbf{(i)}\\\\y=f(x)= \dfrac{x}{4+x}\\\\\\\dfrac{dy}{dx}= \lim \limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}\\\\\\~~~~~=\lim \limits_{h \to 0}\dfrac{\dfrac{x+h}{4+x+h}-\dfrac{x}{4+x}}{h}\\\\\\~~~~~~=\lim \limits_{h \to 0}\dfrac{\tfrac{(x+h)(4+x)-x(4+x+h)}{(4+x+h)(4+x)}}{h}\\\\\\~~~~~~= \lim \limits_{ h \to 0} \dfrac{4x+x^2 +4h+hx -4x-x^2 -hx}{h(4+x+h)(4+x)}\\\\\\~~~~~~=\lim \limits_{ h \to 0}\dfrac{4h}{h(4+x+h)(4+x)}\\\\\\~~~~~~=\lim \limits_{h \to 0} \dfrac{4}{(4+x+h)(4+x)}\\[/tex]
[tex]~~~~~~~~= \dfrac{ 4}{(4+x+0)(4+x)}\\\\\\~~~~~~~~= \dfrac{4}{(4+x)^2}[/tex]
[tex]\textbf{(ii)}\\\\\text{Given that,}~ y = f(x) = \sin 2x\\\\\\\dfrac{dy}{dx} = \lim \limits_{h \to 0} \dfrac{f(x+h) - f(x) }{h}\\\\\\~~~~~~=\lim \limits_{h \to 0} \dfrac{\sin(2x+2h)-\sin 2x}{h}\\\\\\~~~~~~=\lim \limits_{h \to 0} \dfrac{2 \sin \left( \dfrac{2x +2h -2x}{2} \right) \cos \left(\dfrac{2x+2h+2x}{2} \right)}{h}\\\\\\~~~~~~=\lim \limits_{h \to 0} \dfrac{2 \sin \left( \dfrac {2h}2 \right) \cos \left(\dfrac{ 4x+2h} 2\right)}{h}\\\\\\[/tex]
[tex]~~~~~~=\lim \limits_{h \to 0} \dfrac{2 \sin h \cdot \cos \left[ \dfrac{2(2x+h)}{2}\right] }{h}\\\\\\~~~~~~=\lim \limits_{h \to 0} \dfrac{2 \sin h \cdot \cos(2x+h) }{h}\\\\\\~~~~~~=2\left(\lim \limits_{h \to 0} \dfrac{\sin h}{ h} \right) \cdot \lim \limits_{h \to 0} \cos (2x+h)\\\\\\~~~~~~=2\cdot 1 \cdot \cos(2x+ 0)\\\\\\~~~~~~=2 \cos 2x[/tex]
