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

Question

49085 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-10T20:05:42

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

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 $d/dx$

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

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

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