A calculus problem by Jeremy marquardt

first, multiply both sides by dx

$2xdx=\frac{y^{ln(x)}}{1+(x^{ln(y)})^2}ln(x^{\frac{y'}{y}}y^{\frac{1}{x}})dx$

Then, separate the $ln(x^{\frac{y'}{y}}y^{\frac{1}{x}})$ and bring out the exponents, which will yield

$2xdx = \frac{y^{ln(x)}}{1+(x^{ln(y)})^2}(\frac{ln(x)}{y}\frac{dy}{dx}+\frac{ln(y)}{x})dx$ (notice how we changed y' to $\frac{dy}{dx}$ )

Next, distribute the dx across $(\frac{ln(x)}{y}\frac{dy}{dx}+\frac{ln(y)}{x})$

This yields $2xdx = \frac{y^{ln(x)}}{1+(x^{ln(y)})^2}(\frac{ln(x)}{y}dy+\frac{ln(y)}{x}dx)$

notice that $(\frac{ln(x)}{y}dy+\frac{ln(y)}{x}dx)$ equals $d(ln(x)ln(y))$ go ahead and make the transfer

we now have $2xdx = \frac{y^{ln(x)}}{1+(x^{ln(y)})^2}d(ln(x)ln(y))$

now notice that $x^{ln(y)}=y^{ln(x)}=e^{ln(x)ln(y)}$

if we substitute z for ln(x)ln(y), that gives us $2xdx = \frac{e^{z}}{1+(e^z)^2}dz$

now we can integrate this quite easily, and our answer is $x^2 = arctan(y^{ln(x)})$ and solving for y yields $tan(x^2)^{(\frac{1}{ln(x)})}=y$ we said at the beginning of the problem that the constant of integration would be zero, so we don't have to worry about it.

putting f(x) in the configuration of the limit we get $2\frac{ln(x)}{ln(arctan(x))}$

using L'Hopital's rule, $2\frac{(x^2+1)arctan(x)}{x}$

applying it again, $2\frac{2xarctan(x)+1}{1}$

which when evaluated with x=0 equals 2!