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

Question

1907 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-01T14:32:12

#y=ln(x/(x^2+1))#

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

Now, if #g(x)=ln(h(x))#, #g'(x)=(h'(x))/(h(x))#, which means that...

#(dy)/(dx)=x^-1-(2x)/(x^2+1)#

So far so good... Now from here, what we get is...

#(d^2y)/(dx^2)=-x^(-2)-((x^2+1)*2-(2x)^2)/((x^2+1)^2)#

Thanks to the quotient rule.

If we simplify the above we get...

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

#(d^2y)/(dx^2)=-1/(x^2)-(-2x^2+2)/(x^2+1)^2#

#(d^2y)/(dx^2)=-1/(x^2)-(-2(x^2-1))/(x^2+1)^2#

#(d^2y)/(dx^2)=-1/(x^2)+(2(x^2-1))/(x^2+1)^2#