Let f(x)=(x^3-2/x)^-6. Find f'(x)=

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Find the derivative of `f(x)=(x^3-2/x)^(-6)` :

We apply the general power rule (derived from the chain rule) `d/(dx)u^n=n u^(n-1)(du)/(dx)` assuming u is a differentiable function of x.

Let `u=x^3-2/x` so `(du)/(dx)=3x^2+2/x^2` .

So `f'(x)=-6(x^3-2/x)^(-7)(3x^2+2/x^2)`

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