Question

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

Answer 1

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(u)`. As the chain rule states that:

`(dz)/(du)*(du)/(d(xy))=(dz)/(d(xy))`, then

`(dz)/(dx)=(1/u)*(2x+2y)=(1/(x^2+y^2))(2x+2y)=(2x+2y)/(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:

`(deltaz)/(deltax)=(1/u)*(2x)=(2x)/(x^2+y^2)`

`(deltaz)/(deltay)=(1/u)*(2y)=(2y)/(x^2+y^2)`