What is the derivative of ln(x^2 + y^2)?

[ace123]

Find y' if y= ln( x^2 + y^2)

I thought this was just a regular natural log derivative combined with the chain rule.
So what I got was (2x + 2y)/ (x^2 + y^2)

But this wasn't the correct answer. So could someone help point out my mistake.
Thank you.

 

[Vid]

This is implicit differentiation. Since you have y = f(x,y) instead of y = f(x), you have to assume y does equal to some function of x then take derivatives on both sides and solve for y'. Since we made the assumption y can be written as some function of x, D(y^2) = 2yy'. Now, you have to solve for y' just in terms of x and y to get the final answer.

Forgot to mention y' can be in terms of x AND y since we don't know what y is there's no way to remove it from the expression without solving for y in the original equation which is what we were trying to avoid in the first place.

 

[ace123]

Oh okay thank you