Question

Using the limit definition, how do you find the derivative of `f(x)=3(x^(-2))`?

Answer 1

`f'(x) = -6x^(-3)`

Explanation:

According to the limit definition, the derivative of `f(x)` is

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

In your case, this means

`f'(x) = lim_(h -> 0) (3(x+h)^(-2) - 3x^(-2)) /h`

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

`= lim_(h -> 0) ((3x^2 - 3(x+h)^2)/((x+h)^2 *x^2)) /h`

`= lim_(h -> 0) (3x^2 - 3(x+h)^2)/(h(x+h)^2 *x^2)`

... use `(a+b)^2 = a^2 + 2ab + b^2`...

`= lim_(h -> 0) (3x^2 - 3(x^2 + 2xh + h^2))/(h(x+h)^2 *x^2)`

`= lim_(h -> 0) (cancel(3x^2) - cancel(3x^2) - 6xh - 3h^2)/(h(x+h)^2 *x^2)`

`= lim_(h -> 0) (h( - 6x - 3h))/(h(x+h)^2 *x^2)`

... cancel `h`....

`= lim_(h -> 0) ( - 6x - 3h)/((x+h)^2 *x^2)`

... at this point you can apply the `lim`, so plug `h=0`:

`= ( - 6x - 0)/((x+0)^2 *x^2)`

`= ( - 6x )/(x^4)`

`= ( - 6 )/(x^3)`

`= -6x^(-3)`

Thus, your derivative is

`f'(x) = -6x^(-3)`