Derivative of $ y = \ln(x^2 + y^2)$? Why is $d(y^2)/dx = 0$?

Question

The answer is

$y' = 2x/(x^2+y^2-2y)$

or

$y' = 2x/(x^2+y^2)$

At wolfram it says $d(y^2)/dx = 0$?

http://www4c.wolframalpha.com/Calculate/MSP/MSP44041daf8be8cc1i97i900005a2dbcggic267i1c?MSPStoreType=image/png&s=32&w=560&h=716

Answer 1

If we have:

$$f(x, y) = \ln(x^2 + y^2)$$

Note that the $y$ term is missing here, and using the chain rule with $\ln$, we get:

$$d/dx ~(\ln(x^2 + y^2)) = \dfrac{1}{x^2 + y^2}~d/dx~ (x^2 + y^2) = \dfrac{1}{x^2 + y^2}(2x+0) =\dfrac{2x}{x^2 + y^2}$$

This is your second result, but I do not believe this is what you wanted.

Update

If we are doing implicit differentiation of your original expression, we have:

$$dy/dx = \dfrac{1}{x^2+y^2}(2x + 2y ~ dy/dx)$$

Solving for $dy/dx$ yields:

$$\dfrac{dy}{dx} = \dfrac{2x}{x^2+y(y-2)}$$

This is your first result. If you wanted to see this in WA, try this.