What is the derivative for x^2 + y^2 + ln(2) = xy?

[antinerd]

You are close, but we want the derivative of y^2 wrt x, so we need to apply the chain rule. Do you know the chain rule d/dx [u^2]=2u*du/dx

How can I use the chain rule here?

 

[abelian jeff]

How can I use the chain rule here?

See post #25.

 

[bomba923]

Good point! I guess we can assume x and y are complex...

All real numbers are complex; perhaps you meant, "assume x or y have nonzero imaginary coefficients?"

I don't think it affects the implicit differentiation does it?

You can still implicitly differentiate 'y' with respect to 'x' in x^2 + y^2 + \ln 2 = xy
to obtain
y' = \frac{{2x - y}}{{x - 2y}}
with simple poles spanning the (complex) line x = 2y as expected

Yeah, well, that's the question that's written down...

Hmm... I guess it doesn't really matter. Does it even affect what I will get as a derivative?

There is not much difference if you set your domain to be \mathbb{C}^2 - \mathbb{R}^2;
otherwise your original equation cannot solve your derivative (i.e, with domain \mathbb{R}^2)

 

[antinerd]

OK so, guys, I just want to thank you for the help so far :)

This is getting the dx/dy with respect to x and y, correct?

x^2 + y^2 + ln2 = xy

(d/dx) x2 + (d/dx) y^2 + (d/dx) ln2 = (d/dx) xy

2x + 2y dy/dx + 0 = y + x dy/dx

Is this correct so far? If not, what am I doing wrong? I would appreciate corrections :)

 

[matt grime]

You're writing things like (dy/dx)y^2 when you should be writing (d/dx)y^2.

The first is multipying a function dy/dx by another function y^2, the second is applying the derivative operator d/dx to the function y^2, which is what you should be doing.

 

[Dick]

That is correct. Now solve for dy/dx.

 

[antinerd]

That is correct. Now solve for dy/dx.

That's where I need the help... :(

Is it now just algebraic manipulation, or do I have to do something with the terms:

2y (dy/dx) and x (dy/dx)

?

 

[Dick]

It is just algebraic manipulation. The calculus is over. Move both terms to one side of the equation and factor out dy/dx.

 

[antinerd]

You're the man, Dick!

Thanks to everyone.

Final answer:

dy/dx = (-x+1) / y

PLEASE TELL ME THIS IS CORRECT! THEN I CAN BLOW THROUGH THE REST OF 'EM!

 

[learningphysics]

You're the man, Dick!

Thanks to everyone.

Final answer:

dy/dx = (-x+1) / y

PLEASE TELL ME THIS IS CORRECT! THEN I CAN BLOW THROUGH THE REST OF 'EM!

I'm getting dy/dx = (y-2x)/(2y-x)

check your algebra.

 

[antinerd]

I'm getting dy/dx = (y-2x)/(2y-x)

check your algebra.

Wait I got

(2x-y)/(x-2y) = dy/dx

...

 

[antinerd]

Because isn't it:

(2x-y) = dy/dx (x-2y)

and then you divide to get dy/dx = (2x-y)/(x-2y)

 

[learningphysics]

Because isn't it:

(2x-y) = dy/dx (x-2y)

and then you divide to get dy/dx = (2x-y)/(x-2y)

same thing. (if I multiply the numerator and denominator by -1, I'll get it that way)...

I took the dy/dx to the left side when solving, you took it to the right... it's the same answer.

 

[ZioX]

Pffsh. Equations don't have derivatives. Implicit functions do, however.

 

[Saladsamurai]

Pffsh. Equations don't have derivatives. Implicit functions do, however.

Your mom has derivatives.

 

[bomba923]

Don't be confused by the title of this thread

Pffsh. Equations don't have derivatives. Implicit functions do, however.

Rest assured no one was trying to "differentiate an equation"