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If y = ln (x^2 + y^2), find dy/dx
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[d/dx] y = [d/dx] ln (x^2 + y^2) dy/dx = [1/(x^2 + y^2)] (2x) + (2y)(dy/dx)
@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)?
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