What is y' if y= ln [(x^2)+(y^2)]?

Question

Supposedly, the answer isn't (2x+2y)/(x^2+y^2) which is what I got...
Please explain

Answer 1

You're pretty close. You missed a step in the chain rule when differentiating x^2 + y^2: the correct derivative for this (wrt x) is 2x + 2y y'.

More fully, implicit differentiation gives
y' = 1/(x^2 + y^2) . d/dx(x^2 + y^2) (chain rule)
= 1/(x^2 + y^2) [2x + 2y dy/dx] (chain rule)
= (2x + 2y y') / (x^2 + y^2)

In general, in implicit differentiation, you should expect y' to appear on both sides of the equation.

To isolate y', we have to proceed as follows:
(x^2 + y^2) y' = 2x + 2y y'
=> (x^2 + y^2 - 2y) y' = 2x
=> y' = 2x / (x^2 + y^2 - 2y).

Answer 2

You have to use the chain rule and implicit differentiation.

dy/dx = (1/(x^2 + y^2)) * (2x + 2y*dy/dx)

dy/dx = 2x/(x^2 + y^2) + (2y*dy/dx)/(x^2 + y^2)

dy/dx - (dy/dx)*(2y)/(x^2 + y^2) = 2x/(x^2 + y^2)

(dy/dx)(1 - 2y/(x^2 + y^2) = 2x/(x^2 + y^2)

dy/dx = 2x/((x^2 + y^2) * (1 - 2y/(x^2 + y^2))