How do you find the first and second derivative of #(ln(x^2-1))/x^2#?

1 Answer
Aug 22, 2016

Answer:

#y'=[2{x^2-(x^2-1)ln(x^2-1)}]/(x^3(x^2-1)#.

#y''=2/x^2{(3ln(x^2-1))/x^2-(x^2+1)/(x^2-1)^2}#.

Explanation:

Let #y=(ln(x^2-1))/x^2........(1)#

#:. x^2y=ln(x^2-1)=ln{(x+1)(x-1)}#

#x^2y=ln(x+1)+ln(x-1)#.

Diff. ing both sides w.r.t. #x#, we get,

#x^2*y'+y(x^2)'=(ln(x+1))'+(ln(x-1))'#.

#:. x^2y'+y(2x)={1/(x+1)}(x+1)'+{1/(x-1)}(x-1)'#.

#:. x^2y'+2xy=1/(x+1)+1/(x-1)=(2x)/(x^2-1)..........(2)#

#:. y'=1/x^2[2x{1/(x^2-1)-y}]#, or, using #(1)# for #y#,

#y'=2/x{1/(x^2-1)-(ln(x^2-1))/x^2}#

Hence, #y'=[2{x^2-(x^2-1)ln(x^2-1)}]/(x^3(x^2-1)#.

For #y''#, we rewrite #(2)#, in the following more useful form :

#x^2y'+2xy=1/(x+1)+1/(x-1)#.

Diff.ing, both sides w.r.t. #x#, we get,

#x^2y''+y'(2x)+2(xy'+y)=-1/(x+1)^2-1/(x-1)^2#

#:. x^2y''+4xy'+2y=-{((x-1)^2+(x+1)^2)/((x+1)^2(x-1)^2)}#

#:. x^2y''+4xy'+2y=-(2(x^2+1))/(x^2-1)^2#

Here, #xy'=2{1/(x^2-1)-y}#

#:. x^2y''+4*2{1/(x^2-1)-y}+2y=-(2(x^2+1))/(x^2-1)^2#

#:. x^2y''+8/(x^2-1)-8y+2y=-(2(x^2+1))/(x^2-1)^2#

#:. x^2y''-6y= -(2(x^2+1))/(x^2-1)^2#

#:. x^2y''=6y-(2(x^2+1))/(x^2-1)^2=2{3y-(x^2+1)/(x^2-1)^2}#

#=2{(3ln(x^2-1))/x^2-(x^2+1)/(x^2-1)^2}#

Finally, #y''=2/x^2{(3ln(x^2-1))/x^2-(x^2+1)/(x^2-1)^2}#.

Enjoy Maths.!