(i) Assume, $f{\text{ }}\left( x \right){\text{ }} = {\text{ }}{x^3}\;–{\text{ }}27$.

Using first principle we have,

${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{{\text{h}} \to 0} \frac{{{\text{f}}({\text{x}} + {\text{h}}) – {\text{f}}({\text{x}})}}{{\text{h}}}$

${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{({\text{x}} + {\text{h}})}^3} – 27} \right] – \left( {{{\text{x}}^3} – 27} \right)}}{{\text{h}}}$

${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{h \to 0} \frac{{{{\text{x}}^3} + {{\text{h}}^3} + 3{{\text{x}}^2}\;{\text{h}} + 3{\text{x}}{{\text{h}}^2} – {{\text{x}}^3}}}{{\;{\text{h}}}}$

${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{h \to 0} \frac{{{{\text{h}}^3} + 3{{\text{x}}^2}\;{\text{h}} + 3{\text{x}}{{\text{h}}^2}}}{{\;{\text{h}}}}$

${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{{\text{h}} \to 0} \left( {\;{{\text{h}}^2} + 3{{\text{x}}^2} + 3{\text{xh}}} \right)$

${{\text{f}}^\prime }({\text{x}}) = 0 + 3{{\text{x}}^2}$

${{\text{f}}^\prime }({\text{x}}) = 3{{\text{x}}^2}$

(ii) Assume, f (x) = (x – 1) (x – 2)

Using first principle we have,

${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{{\text{h}} \to 0} \frac{{{\text{f}}({\text{x}} + {\text{h}}) – {\text{f}}({\text{x}})}}{{\text{h}}}$

${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{{\text{h}} \to 0} \frac{{({\text{x}} + {\text{h}} – 1)({\text{x}} + {\text{h}} – 2) – ({\text{x}} – 1)({\text{x}} – 2)}}{{\text{h}}}$

${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{ = {\text{h}} \to 0} \frac{{\left( {{{\text{x}}^2} + {\text{hx}} – 2{\text{x}} + {\text{hx}} + {{\text{h}}^2} – 2\;{\text{h}} – {\text{x}} – {\text{h}} + 2} \right) – \left( {{{\text{x}}^2} – 2{\text{x}} – {\text{x}} + 2} \right)}}{{\text{h}}}$

${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{{\text{h}} \to 0} \frac{{{\text{hx}} + {\text{hx}} + {{\text{h}}^2} – 2\;{\text{h}} – {\text{h}}}}{{\text{h}}}$

${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{{\text{h}} \to 0} (\;{\text{h}} + 2{\text{x}} – 3)$

${{\text{f}}^\prime }({\text{x}}) = 0 + 2{\text{x}} – 3$

${{\text{f}}^\prime }({\text{x}}) = 2{\text{x}} – 3$