( i ) Suppose, $f(x) =  – x$.

This implies, $f(x + h) =  – (x + h)$

By the first principle we have,

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

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

$ = \mathop {\lim }\limits_{h \to 0} \frac{{ – x – h + x}}{h}$

$ = \mathop {\lim }\limits_{h \to 0} \frac{{ – h}}{h}$

$ = \mathop {\lim }\limits_{h \to 0} ( – 1)$

$ =  – 1$

(ii) Suppose, ${\text{f}}({\text{x}}) = {( – {\text{x}})^{ – 1}} = \frac{1}{{ – {\text{x}}}} = \frac{{ – 1}}{{\text{x}}}$

This implies, $f(x + h) =  – \frac{1}{{(x + h)}}$

By the first principle we have,

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

\[ = \mathop {\lim }\limits_{{\text{h}} \to 0} \frac{1}{{\;{\text{h}}}}\left[ {\frac{{ – 1}}{{{\text{x}} + {\text{h}}}} – \left( {\frac{{ – 1}}{{\text{x}}}} \right)} \right]\]

\[ = \mathop {\lim }\limits_{h \to 0} \frac{1}{{\;{\text{h}}}}\left[ {\frac{{ – 1}}{{{\text{x}} + {\text{h}}}} + \frac{1}{{\text{x}}}} \right]\]

\[ = \mathop {\lim }\limits_{{\text{h}} \to 0} \frac{1}{{\;{\text{h}}}}\left[ {\frac{{ – {\text{x}} + ({\text{x}} + {\text{h}})}}{{{\text{x}}({\text{x}} + {\text{h}})}}} \right]\]

\[ = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ – x + x + h}}{{x(x + h)}}} \right]\]

\[ = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{h}{{x(x + h)}}} \right]\]

\[ = \mathop {\lim }\limits_{h \to 0} \frac{1}{{x(x + h)}}\]

\[ = \frac{1}{{x \cdot x}}\]

\[ = \frac{1}{{{x^2}}}\]