$\displaystyle y^2 = cosh(xy) + log(x^2 + y^2).$
Find $\displaystyle \frac{dy}{dx}.$
This one has got me well knobbed. I don't even know where to start. Maybe root the whole thing? Can someone give me a pointer please?
I did that but then I get an answer with loads of y's and dy/dx's and I don't know how to clean it up...?
EDIT: Ok, I re-did it and got this horrendously disgusting answer? I don't think its right at all? :S
$\displaystyle \frac{dy}{dx}=\frac{ysinh(xy) + \frac{2x}{x^2+y^2}}{2y-xsinh(xy)-\frac{2y}{x^2+y^2}} $