By definition

`f'(x)=lim_(Delta x->0)(f(x+ Delta x)-f(x))/(Delta x)`

So for `f(x)=1/x^2` we have

`f'(x)=lim_(Delta x->0)(1/(x+Delta x)^2-1/x^2)/(Delta x)=lim_(Delta x->0)(1/(x^2+2x Delta x+Delta x^2)-1/x^2)/(Delta x)=`

`lim_(Delta x->0)((x^2-x^2-2x Delta x+Delta x^2)/(x^2(x^2+2x Delta x+Delta x^2)))/(Delta x)=lim_(Delta x->0)(Delta x(-2x+Delta x))/(x^2(x^2+2x Delta x+Delta x^2)Delta x)=`

`lim_(Delta x->0)(-2x+Delta x)/(x^2(x^2+2x Delta x+Delta x^2))=(-2x)/(x^2 cdot x^2)=-2/x^3`

**So your solution is** `f'(x)=-2/x^3`

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