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

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Answer 1 · 2018-04-17T04:40:58

The second derivative is #-(2(3x^2+1))/(x^2-1)^3#.

First, compute the first derivative using the quotient rule:

#color(white)=d/dx[1/(1-x^2)]#

#=(d/dx[1]*(1-x^2)-1*d/dx[1-x^2])/(1-x^2)^2#

#=(0*(1-x^2)-1\*-2x)/(1-x^2)^2#

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

Now the second derivative will be the derivative of this:

#color(white)=d/dx[d/dx[1/(1-x^2)]]#

#=d/dx[(2x)/(1-x^2)^2]#

#=(d/dx[2x]*(1-x^2)^2-2x*d/dx[(1-x^2)^2])/(1-x^2)^4#

#=(2(1-x^2)^2-2x*2(1-x^2)*d/dx[1-x^2])/(1-x^2)^4#

#=(2(1-x^2)^2-2x*2(1-x^2)*-2x)/(1-x^2)^4#

#=(2(1-x^2)^2+8x^2(1-x^2))/(1-x^2)^4#

Factor out #2(1-x^2)#:

#=(2(1-x^2)((1-x^2)+4x^2))/(1-x^2)^4#

#=(2(1-x^2)(1-x^2+4x^2))/(1-x^2)^4#

#=(2(1-x^2)(1+3x^2))/(1-x^2)^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!