Question

How do you use the limit definition to find the derivative of `f(x)=1/x^2`?

Answer 1

The limit definition of the derivative tells us that

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

Since `f(x) = 1/(x^2)` we write

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

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

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

`= lim_(h->0) ((cancel(x^2)cancel(-x^2)-2xh-h^2)/(x^2(x^2+2xh+h^2)))/(h)`

`= lim_(h->0) ((cancel(h)(-2x-h))/(x^2(x^2+2xh+h^2))*1/cancel(h))/(cancel(h/1*1/h)`

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

Substituting `h = 0` gives us

`= (-2x-(0))/(x^2(x^2+2x(0)+(0)^2)) = (-2x)/(x^4) = -2/x^3`

We can check our answer by following the power rule derivatives really quickly:

`f(x) = 1/x^2 = x^(-2) -> f'(x) = -2x^(-3) = -2/(x^3)`