Originally Posted by

**Chris L T521** Yes, but it can get a little more complex...and messy... >_>

$\displaystyle \frac{\,dy}{\,dx}=\frac{1-x+2y}{16y^2-8xy+2}$

Therefore $\displaystyle \frac{d^2y}{dx^2}=\frac{\left(16y^2-8xy+2\right)\left(-1+2\frac{\,dy}{\,dx}\right)-\left(1-x+2y\right)\left(32y\frac{\,dy}{\,dx}-8y-8x\frac{\,dy}{\,dx}\right)}{(16y^2-8xy+2)^2}$

..and then wherever you see $\displaystyle \frac{\,dy}{\,dx}$, plug in $\displaystyle \frac{1-x+2y}{16y^2-8xy+2}$...

Its gonna get messy...

I hope this helps.

--Chris