How to find the derivative of ln ( x 2 + y 2 ) ?

Question

How to find the derivative of             ln              (                  x          2                +                  y          2                )            ?

Desiree Cantrell · Accepted Answer

Implicit differentiation and partial differentiation are the two possible directions here.
For implicit differentiation, we have both variables (

x

and

y

) into the derivative at once, while for partial differentiation we work with each one separately.
Implicitly differetiating, then, we must resort to chain rule, by naming

u = x^{2} + y^{2}

and, therefore, considering our original funcion

z = \ln (u)

. As the chain rule states that:

\frac{d z}{d u} \cdot \frac{d u}{d (x y)} = \frac{d z}{d (x y)}

, then

\frac{d z}{d x} = (\frac{1}{u}) \cdot (2 x + 2 y) = (\frac{1}{x^{2} + y^{2}}) (2 x + 2 y) = \frac{2 x + 2 y}{x^{2} + y^{2}}

Now, going to partial differentiation: we keep the chain rule logic, but in the end, we proceed differently, differentiating only one of the two variables, as follows:

\frac{δ z}{δ x} = (\frac{1}{u}) \cdot (2 x) = \frac{2 x}{x^{2} + y^{2}}

\frac{δ z}{δ y} = (\frac{1}{u}) \cdot (2 y) = \frac{2 y}{x^{2} + y^{2}}

How to find the derivative of ln(x^2+y^2)?

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