Question

How do you use the limit definition to find the derivative of `f(x)=(2-x)/(3x+1)`?

Answer 1

`d/(dx) ((2-x)/(3x+1)) = -7/(3x+1)^2`

Explanation:

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

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

Calculate:

`Deltaf = (2-(x+Deltax))/(3(x+Deltax)+1) - (2-x)/(3x+1)`

`Deltaf =( (2-x-Deltax)(3x+1) -(3x+3Deltax+1) (2-x))/((3x+3Deltax+1)(3x+1))`

`Deltaf =( cancel((2-x)(3x+1))-Deltax(3x+1) -cancel((3x+1) (2-x))-3Deltax(2-x))/((3x+3Deltax+1)(3x+1))`

`Deltaf = ( -Deltax(3x+1+6-3x))/((3x+3Deltax+1)(3x+1)) = -(7Deltax)/((3x+Deltax+1)(3x+1))`

Divide by `Deltax`:

`(Deltaf)/(Deltax) = -7/((3x+Deltax+1)(3x+1))`

Passing to the limit:

`lim_(Deltax->0) (Deltaf)/(Deltax) = lim_(Deltax->0) -7/((3x+Deltax+1)(3x+1)) = - 7/((3x+1)^2)`