Lesson 4

# Partial Differentiation

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$(\dfrac{\partial z}{\partial x})^2 + (\dfrac{\partial z}{\partial y})^2 \;=\; 4\sqrt{u^2+v^2} \bigg[ (\dfrac{\partial z}{\partial u})^2 + (\dfrac{\partial z}{\partial v})^2 \bigg] \ \; \ \; \ \; \$

prove that

$x^2 \dfrac{\partial u}{\partial x} + y^2 \dfrac{\partial u}{\partial y} + z^2 \dfrac{\partial u}{\partial z} \;=\; 0$

Prove : $\bigg(\dfrac{\partial u}{\partial x} \bigg)^2 + \bigg(\dfrac{\partial u}{\partial y} \bigg)^2 + \bigg(\dfrac{\partial u}{\partial z} \bigg)^2 \;=\; 2 \bigg( x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} + z \dfrac{\partial u}{\partial z} \bigg) \\ \; \\$

$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) \ \; \$
then prove that $\dfrac{\partial^2u}{\partial \theta^2} + \dfrac{\partial^2u}{\partial \phi^2} \;=\; 4xy \dfrac{\partial^2 u}{\partial x \partial y}$
$x \dfrac{\partial u}{\partial x} + y \dfrac{\partial u}{\partial y} + z \dfrac{\partial u}{\partial z}$ where $u \;=\; \dfrac{x^3y^3z^3}{x^3+y^3+z^3}$