Question

If y = ln (x^2 + y^2), find dy/dx

Answer 1

[d/dx] y = [d/dx] ln (x^2 + y^2)

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

Answer 2

@petewe Wouldn't it be (2x + 2y (dy/dx) ) / (x^2 + y^2) since you multiply 1/(x^2 + y^2) by the quantity of the derivative of (x^2 + y^2)?