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

[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"