# Mixed differentiation?

• Jun 9th 2011, 12:33 PM
TeaWithoutMussolini
Mixed 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 PM
TheEmptySet
Quote:

Originally Posted by TeaWithoutMussolini
$\displaystyle y^2 = cosh(xy) + log(x^2 + y^2).$

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?

Use implicit differentiation and the chain rule.

$\displaystyle 2y\cdot \frac{d}{dx}y=\sinh(xy)\cdot \frac{d}{dx}(xy)+\frac{1}{x^2+y^2}\cdot \frac{d}{dx}\left(x^2+y^2 \right)$

Can you finish from here
• Jun 9th 2011, 01:49 PM
TeaWithoutMussolini
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 PM
TheEmptySet
Quote:

Originally Posted by TeaWithoutMussolini
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?

No All I did was use the chain rule you need to expand all of these $\displaystyle \frac{d}{dx}(xy)$ and $\displaystyle \frac{d}{dx}(x^2+y^2)$ and then collect all of the $\displaystyle y'$'s

This is just the 1st step.
• Jun 9th 2011, 01:55 PM
TeaWithoutMussolini
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 PM
TheEmptySet
Quote:

Originally Posted by TeaWithoutMussolini
Can y be treated as a constant? I'm confused? Would the next step to do the same thing again on the 'new' parts?

y is not a constant y is a function of x. $\displaystyle y=f(x)$ Maybe this will help consider

$\displaystyle \frac{d}{dx}y^2=\frac{d}{dx}[f(x)]^2=\underbrace{2f(x)f'(x)}_{\text{ Chain Rule }}=2yy'$ this is the idea of implicit differentiation!
• Jun 9th 2011, 02:31 PM
TeaWithoutMussolini
So is the $\displaystyle \frac{d}{dx}(xy)$ becomes $\displaystyle x\frac{dy}{dx} + y?$
• Jun 9th 2011, 03:07 PM
TheEmptySet
Quote:

Originally Posted by TeaWithoutMussolini
So is the $\displaystyle \frac{d}{dx}(xy)$ becomes $\displaystyle x\frac{dy}{dx} + y?$

yes!
• Jun 9th 2011, 08:19 PM
TeaWithoutMussolini
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

$\displaystyle \frac{dy}{dx}=\frac{ysinh(xy) + \frac{2x}{x^2+y^2}}{2y-xsinh(xy)-\frac{2y}{x^2+y^2}}$