Question

How do you find the derivative of `f(x)=1/x^2` using the limit process?

Answer 1

`f'(x)=-2/x^3`

Explanation:

By definition of the derivative `f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h`
So with `f(x) = 1/x^2` we have;

`f'(x)=lim_(h rarr 0) ( 1/(x+h)^2 -1/x^2 ) / h`
`:. f'(x)=lim_(h rarr 0)1/h*(x^2-(x+h)^2)/((x+h)^2x^2)`
`:. f'(x)=lim_(h rarr 0) (x^2-(x^2+2hx+h^2))/(h(x+h)^2x^2)`
`:. f'(x)=lim_(h rarr 0) (x^2-x^2-2hx-h^2)/(h(x+h)^2x^2)`
`:. f'(x)=lim_(h rarr 0) (-2x-h)/((x+h)^2x^2)`
`:. f'(x)=(-2x)/((x)^2x^2)`
`:. f'(x)=-2/x^3`