How do you use the limit definition to find the derivative of f(x)=(2-x)/(3x+1)f(x)=2x3x+1?

1 Answer
Jan 21, 2017

d/(dx) ((2-x)/(3x+1)) = -7/(3x+1)^2ddx(2x3x+1)=7(3x+1)2

Explanation:

By definition the derivative of f(x)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)