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What is the second order derivative of ##y = 1 / (1-x^2)##?
##y''=-2(3x^2+1)/(x^2-1)^3##
##y=1/(1-x^2)=-1/(x^2-1)##
Using the and linearity of the differantation we obtain:
##y'=[k*(h(x))/g(x)]^'=k[(h(x))/g(x)]^'=k[(h'(x)*g(x)-g'(x)*h(x))/g(x)^2]##
where:
##k=-1## ##h(x)=1## ##g(x)=x^2-1##
##:.y'=-1*[(0*(x^2-1)-(2x+0)*1)/(x^2-1)^2]=##
##=-1*[(-2x)/(x^2-1)]=(2x)/(x^2-1)^2##
Now we could apply again the and linearity of the differantation to compute the second order derivative
##y''=(d^2y)/dx^2=d/dxy'=##
##=2[(1*(x^2-1)^2-2(x^2-1)*(2x-0)*x)/[(x^2-1)^2]^2]=## ##=2[((x^2-1)^2-4x^2(x^2-1))/(x^2-1)^4]=## ##=2color(magenta)cancel((x^2-1))(x^2-1-4x^2)/(x^2-1)^color(magenta)(cancel(4)^3)=-2(3x^2+1)/(x^2-1)^3##