First, compute the first derivative using the quotient rule:
color(white)=d/dx[1/(1-x^2)]=ddx[11−x2]
=(d/dx[1]*(1-x^2)-1*d/dx[1-x^2])/(1-x^2)^2=ddx[1]⋅(1−x2)−1⋅ddx[1−x2](1−x2)2
=(0*(1-x^2)-1*-2x)/(1-x^2)^2=0⋅(1−x2)−1⋅−2x(1−x2)2
=(2x)/(1-x^2)^2=2x(1−x2)2
Now the second derivative will be the derivative of this:
color(white)=d/dx[d/dx[1/(1-x^2)]]=ddx[ddx[11−x2]]
=d/dx[(2x)/(1-x^2)^2]=ddx[2x(1−x2)2]
=(d/dx[2x]*(1-x^2)^2-2x*d/dx[(1-x^2)^2])/(1-x^2)^4=ddx[2x]⋅(1−x2)2−2x⋅ddx[(1−x2)2](1−x2)4
Chain rule:
=(2(1-x^2)^2-2x*2(1-x^2)*d/dx[1-x^2])/(1-x^2)^4=2(1−x2)2−2x⋅2(1−x2)⋅ddx[1−x2](1−x2)4
=(2(1-x^2)^2-2x*2(1-x^2)*-2x)/(1-x^2)^4=2(1−x2)2−2x⋅2(1−x2)⋅−2x(1−x2)4
=(2(1-x^2)^2+8x^2(1-x^2))/(1-x^2)^4=2(1−x2)2+8x2(1−x2)(1−x2)4
Factor out 2(1-x^2)2(1−x2):
=(2(1-x^2)((1-x^2)+4x^2))/(1-x^2)^4=2(1−x2)((1−x2)+4x2)(1−x2)4
=(2(1-x^2)(1-x^2+4x^2))/(1-x^2)^4=2(1−x2)(1−x2+4x2)(1−x2)4
=(2(1-x^2)(1+3x^2))/(1-x^2)^4=2(1−x2)(1+3x2)(1−x2)4
=(2color(red)cancelcolor(black)((1-x^2))(1+3x^2))/(1-x^2)^(color(red)cancelcolor(black)4^3)
=(2(1+3x^2))/(1-x^2)^3
If you would like to put all the x terms first:
=(2(3x^2+1))/(-x^2+1)^3
=(2(3x^2+1))/(-(x^2-1))^3
=(2(3x^2+1))/(-1*(x^2-1))^3
=(2(3x^2+1))/((-1)^3*(x^2-1)^3)
=(2(3x^2+1))/(-1*(x^2-1)^3)
=-(2(3x^2+1))/(x^2-1)^3
That's the second derivative. Hope this helped!
