Answer:
f'(x) = [tex]-\frac{9}{x^2}[/tex]
Step-by-step explanation:
i) f(x) = 9 / x
ii) f'(x) = [tex]$\lim_{h\to 0} \frac{f(x+h) - f(x)}{h} $ \hspace{0.2cm}[/tex]
[tex]= $\lim_{h\to 0} \frac{\frac{9}{x+h} - \frac{9}{x} }{h} = \hspace{0.1cm} $\lim_{h\to 0} \frac{9x - 9(x+h)}{hx(x+h)} $ \hspace{0.1cm} = \hspace{0.1cm}$\lim_{h\to 0} \frac{-9h}{hx(x+h)} = $\lim_{h\to 0} \frac{-h}{x(x+h)} = \frac{-9}{x^2}$[/tex]
