Answer
$f’\left( x\right) =\dfrac {2}{x^{3}}$
$f’\left( 3\right) =\dfrac {2}{27}$
$f’\left( 4\right) =\dfrac {1}{32}$
Work Step by Step
$f\left( x\right) =\dfrac {-1}{x^{2}}=-x^{-2}\Rightarrow f'\left( x\right) =\left( -1\right) \times \left( -2\right) \times x^{-2-1}=2\times x^{-3}=\dfrac {2}{x^{3}}$
$f’\left( 3\right) =\dfrac {2}{x^{3}}=\dfrac {2}{3^{3}}=\dfrac {2}{27}$
$f’\left( 4\right) =\dfrac {2}{x^{3}}=\dfrac {2}{4^{3}}=\dfrac {2}{64}=\dfrac {1}{32}$
