Allgemein:
y=x^n
y' =n *x^{n-1}
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f(x)= 5/x^2 = 5 *x^{-2}
f ' (x)= 5*(-2) x^{-3}
f '(x)= -10 x^{-3}
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Es wird summandweise differenziert.
f(x)= x^2+4/x =x^2 + 4*x^{-1}
f '(x)= x^2 + (-1) *4*x^{-2}
f '(x) = 2x - 4 x^{-2}
f '(x) = 2x - 4 /x^2