then prove that $ \dfrac{\partial^2u}{\partial \theta^2} + \dfrac{\partial^2u}{\partial \phi^2} \;=\; 4xy \dfrac{\partial^2 u}{\partial x \partial y} $

Lesson 4

# Partial Differentiation

#### Euler theorem

$x \dfrac{\partial u}{\partial x} + y \dfrac{\partial u}{\partial y} + z \dfrac{\partial u}{\partial z} $ where $ u \;=\; \dfrac{x^3y^3z^3}{x^3+y^3+z^3} $

$x^2 \dfrac{\partial^2 u}{\partial x^2}+ 2xy\dfrac{\partial^2 u}{\partial x \partial y}+ y^2\dfrac{\partial^2 u}{\partial y^2}+ x\dfrac{\partial u}{\partial x}+ y \dfrac{\partial u}{\partial y} \ \; \ \; \ For \; u= e^{x+y} \; + \; log(x^3+y^3-x^2y-xy^2) \ \; \ $