$\displaystyle y^2 = cosh(xy) + log(x^2 + y^2).$

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?

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- Jun 9th 2011, 12:33 PMTeaWithoutMussoliniMixed differentiation?
$\displaystyle y^2 = cosh(xy) + log(x^2 + y^2).$

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? - Jun 9th 2011, 01:41 PMTheEmptySet
- Jun 9th 2011, 01:49 PMTeaWithoutMussolini
Thanks, I think I can handle the operations but I don't know what kind of form I would be expected to leave it in? I mean is that not done? Just divide by 2y?

- Jun 9th 2011, 01:52 PMTheEmptySet
- Jun 9th 2011, 01:55 PMTeaWithoutMussolini
Can y be treated as a constant? I'm confused? Would the next step to do the same thing again on the 'new' parts?

- Jun 9th 2011, 01:59 PMTheEmptySet
- Jun 9th 2011, 02:31 PMTeaWithoutMussolini
So is the $\displaystyle \frac{d}{dx}(xy)$ becomes $\displaystyle x\frac{dy}{dx} + y?$

- Jun 9th 2011, 03:07 PMTheEmptySet
- Jun 9th 2011, 08:19 PMTeaWithoutMussolini
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}} $