Question

How do you use the limit definition to find the derivative of `f(x)=1/(4x-3)`?

Answer 1

`f'(x) = lim_(Deltax->0) (Deltaf)/(Deltax) =(-4)/((4x-3)^2)`

Explanation:

The definition of the derivative of `f(x)` is:

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

Let us calculate the increment of `f(x)`:

`Deltaf = 1/(4(x+Deltax)-3) -1/(4x-3)`

`Deltaf = (4x-3-4x-4Deltax+3)/((4x+4Deltax-3)(4x-3))=(-4Deltax)/((4x+4Deltax-3)(4x-3))`

So:

`(Deltaf)/(Deltax) =(-4)/((4x+4Deltax-3)(4x-3))`

and

`lim_(Deltax->0) (Deltaf)/(Deltax) =(-4)/((4x-3)^2)`