1. ## partial differentiation problem

Hi

can anyone help me partially differentiate

u = (-2xyz)/(x^2 + y^2)^2

Partial du/dx = ?

and

v = (z(x^2-y^2))/(x^2+y^2)^2

Partial dv/dx = ?

Any ideas for the second partial derivatives?

Any help would be greatly appreciated

Many thanks

zentus

2. Hello, zentus!

Partial derivatives

$\displaystyle u \:= \:\frac{-2xyz}{(x^2 + y^2)^2}$ . . . Find $\displaystyle \frac{\partial u}{\partial x}$

To differentiate with respect to $\displaystyle x$, treat the other variables as constants.

$\displaystyle \frac{\partial u}{\partial x} \;=\;\frac{(x^2+y^2)\cdot(-2yz) - (-2xyz)\cdot2(x^2+y^2)\cdot2x}{(x^2+y^2)^4} \;=$ .$\displaystyle \frac{-2yz(x^2+y^2)^2 + 8x^2yz(x^2+y^2)}{(x^2+y^2)^4}$

Factor: .$\displaystyle \frac{-2yz(x^2+y^2)\cdot\left[(x^2+y^2) - 4x\right]}{(x^2+y^2)^4}$

Reduce: .$\displaystyle \frac{\partial u}{\partial x} \;=\;\frac{-2yz(x^2+y^2 - 4x)}{(x^2+y^2)^3}$

Now try the second one yourself . . .

Soroban

Thank you!