# How do you find the derivative of ln(x^2+y^2)?

May 16, 2015

There are two possible pathways here: implicit differentiation or partial differentiation.

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 $z = \ln \left(u\right)$. As the chain rule states that:

$\frac{\mathrm{dz}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{d \left(x y\right)} = \frac{\mathrm{dz}}{d \left(x y\right)}$, then

$\frac{\mathrm{dz}}{\mathrm{dx}} = \left(\frac{1}{u}\right) \cdot \left(2 x + 2 y\right) = \left(\frac{1}{{x}^{2} + {y}^{2}}\right) \left(2 x + 2 y\right) = \frac{2 x + 2 y}{{x}^{2} + {y}^{2}}$

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:

$\frac{\delta z}{\delta x} = \left(\frac{1}{u}\right) \cdot \left(2 x\right) = \frac{2 x}{{x}^{2} + {y}^{2}}$

$\frac{\delta z}{\delta y} = \left(\frac{1}{u}\right) \cdot \left(2 y\right) = \frac{2 y}{{x}^{2} + {y}^{2}}$