Answer
$f'(x)=-2(x^3-\frac{7}{x})^{-3}(3x^2+\frac{7}{x^2}) $
Work Step by Step
Given that $f(x)=(x^3-\frac{7}{x})^{-2}$
We can write this as $f(x)=(g(x))^{-2}$, where $f(x)$ is the outside function and $g(x)=x^3-\frac{7}{x} = x^3-7x^{-1}$ is the inside function.
To find the answer we need to differentiate the outside function $f(x) $and times it by the derivative of the inside function $g(x)$.
$f(x)=(x^3-\frac{7}{x})^{-2}$
$ f (x)=(g (x))^{-2}$
$f'(x)=-2(g ( x ))^{-2-1}×(g' ( x )) $
$f'(x)=-2(g ( x ))^{-3}×(3x^{3-1} +7x^{-1-1})$
$f'(x)=-2(x^3-\frac{7}{x} )^{-3}×(3x^{2} +7x^{-2}) $
$f'(x)=-2(x^3-\frac{7}{x})^{-3}(3x^2+\frac{7}{x^2}) $
