Question

What is the derivative of `ln(x^2+y^2)`?

Answer 1

It depends on whether we need `d/dx` or `d/dt`

Explanation:

For `d/dt`

`d/dt(ln(x^2+y^2)) = 1/(x^2+y^2) d/dt(x^2+y^2)`

`= 1/(x^2+y^2) * (2xdx/dt + 2y dy/dt)`

For `d/dx`

`d/dx(ln(x^2+y^2)) = 1/(x^2+y^2) d/dx(x^2+y^2)`

`= 1/(x^2+y^2) * (2x+ 2y dy/dx)`