Question

How do you find f'(x) using the definition of a derivative for `f(x)= 1/(x-3)`?

Answer 1

Please see the explanation section below.

Explanation:

`f(x) = 1/(x-3)`

Definition of derivative: `f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h`.

So we have

`f'(x) = lim_(hrarr0) (1/((x+h)-3)-1/(x-3))/h`.

If we try to evaluate the limit by substitution, we get the indeterminate form `0/0`.

We need to rewrite. Our goal is to make the denominator no longer go to `0`. Then we can find the limit using the quotient property of limits.

It is probably not clear to a beginning student what might work, so think about what you could do. Then see if that helps.

The smart thing to do here is to write the numerator as a single fraction.

`lim_(hrarr0) (1/((x+h)-3)-1/(x-3))/h = lim_(hrarr0) ((x-3)/((x-3)(x+h-3))-(x+h-3)/((x-3)(x+h-3)))/h`

`= lim_(hrarr0)(((x-3)-(x+h-3))/((x-3)(x+h-3)))/h`

`= lim_(hrarr0)((-h)/((x-3)(x+h-3)))/h`

If we try substitution, we still get `0/0`, but we can reduce the fraction now.

`= lim_(hrarr0)((-h)/((x-3)(x+h-3)))/(h/1)`

`= lim_(hrarr0)(-h)/((x-3)(x+h-3))*1/h`

`= lim_(hrarr0)(-1)/((x-3)(x+h-3))`

Now the numerator clearly does not approach `0`, so the form will not be indeterminate.

`= (-1)/((x-3)(x+0-3)) = (-1)/(x-3)^2`

That is,

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