# What is the second order derivative of y = 1 / (1-x^2)?

##y”=-2(3x^2+1)/(x^2-1)^3##

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##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##

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