What is the first derivative and second derivative of #(2x^2)/(x^2-1)#?

Question

3299 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-18T04:07:30

#dy/dx = -(4x)/(x^2 - 1)^2 #

#(d^2y)/dx^2 = (12x^2+4)/(x^2-1)^3#

Given:

#(2x^2)/(x^2 - 1)#

Add 0 to the numerator in the form of #- 2 + 2#:

#(2x^2 - 2 + 2)/(x^2 - 1) #

Split into 2 fractions:

#(2x^2 - 2)/(x^2 - 1) + 2/(x^2 - 1) #

The first fraction reduces to 2 and the second fraction can be written as a negative power:

#2 + 2(x^2 - 1)^-1 #

The first derivative is:

#dy/dx = -4x(x^2 - 1)^-2 #

#dy/dx = -(4x)/(x^2 - 1)^2 #

#(d^2y)/dx^2 = (12x^2+4)/(x^2-1)^3#