Question

How do you use the definition of a derivative to find the derivative of `4/x^2`?

Answer 1

Just remember that `4/(x^2)=4x^(-2)` and that `(x^n)'=n x^(n-1)`.

So `(4/(x^2))'=(4x^(-2))'= 4xx(-2)x^(-3)=-8/x^3`

Answer 2

See the explanation section below.

Explanation:

Let `f(x) = 4/x^2`

I will assume that you are using the limit definition in the form

`f'(x) = lim_(hrarr0) (f(x+h)-f(x))/h`

So, we get:

`f'(x) = lim_(hrarr0)(4/(x+h)^2-4/x^2)/h` `" "` (indeterminate form `0/0`)

`= lim_(hrarr0)((4x^2- (4(x+h)^2))/(x^2((x+h)^2)))/(h/1)`

`= lim_(hrarr0)(4x^2-(4x^2+8xh+4h^2))/(x^2(x+h)^2)*1/h`

`= lim_(hrarr0)(-8xh-4h^2)/(x^2(x+h)^2)*1/h`

`= lim_(hrarr0)(-8x-4h)/(x^2(x+h)^2)` `" "` (not an indeterminate form)

`= (-8x)/x^4 = (-8)/x^3`

If you are using `f'(x) = lim_(trarrx)(f(x)-f(t))/(x-t)`

Then the algebra is similar.

`(f(x)-f(t))/(x-t) = (4/x^2-4/t^2)/(x-t)`

`= ((4t^2-4x^2)/(x^2t^2))/((x-t)/1)`

`= (4(t^2-x^2))/(x^2t^2)*1/(x-t)`

`= (-4(x^2-t^2))/(x^2t^2)*1/(x-t)`

`= (-4(x+t)(x-t))/(x^2t^2)*1/(x-t)`

`= (-4(x+t))/(x^2t^2)`

Taking the limit as `trarrx`, we get

`(-4(2x))/(x^2x^2) = (-8)/x^3`