Question

How do you find the derivative of `f(x)=3(x^(−2))` using the limit definition?

Answer 1

Note that

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

The limit definition of the derivative states that the derivative of `f` is

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

So, where `f(x)=3/x^2`, then:

`f'(x)=lim_(hrarr0)(3/(x+h)^2-3/x^2)/h`

Getting a common denominator:

`f'(x)=lim_(hrarr0)((3x^2-3(x+h)^2)/(x^2(x+h)^2))/h`

Rearranging:

`f'(x)=lim_(hrarr0)(3x^2-3(x+h)^2)/(h(x^2)(x+h)^2)`

Expanding the numerator:

`f'(x)=lim_(hrarr0)(3x^2-3(x^2+2hx+h^2))/(h(x^2)(x+h)^2)`

`f'(x)=lim_(hrarr0)(3x^2-3x^2-6hx-3h^2)/(h(x^2)(x+h)^2)`

`f'(x)=lim_(hrarr0)(-6hx-3h^2)/(h(x^2)(x+h)^2)`

We can factor and cancel an `h`:

`f'(x)=lim_(hrarr0)(h(-6x-3h))/(h(x^2)(x+h)^2)`

`f'(x)=lim_(hrarr0)(-6x-3h)/(x^2(x+h)^2)`

We can now evaluate the limit by plugging in `0` for `h`, since there's no longer the issue of `h` in the denominator.

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

`f'(x)=(-6x)/(x^2(x^2))`

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

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