How do you find the second derivative of $ln (\frac{x}{x^{2} + 1})$ $ln(x/(x^2+1))$ ?

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Answer 1 · 2016-11-01T14:32:12

$y = ln (\frac{x}{x^{2} + 1})$ $y=ln(x/(x^2+1))$

$y = ln (x) - ln (x^{2} + 1)$ $y=ln(x)-ln(x^2+1)$

Now, if $g (x) = ln (h (x))$ $g(x)=ln(h(x))$ , $g'(x)=(h'(x))/(h(x))$ , 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...

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

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

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

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

How do you find the second derivative of ln(x/(x^2+1))ln(xx2+1) ln(x/(x^2+1)) ?