partial differentiation problem

**zentus** · November 7th 2007, 01:20 PM

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

**Soroban** · November 7th 2007, 02:37 PM

Hello, zentus!

Partial derivatives

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

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

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

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

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

Now try the second one yourself . . .

**zentus** · November 7th 2007, 10:49 PM

Soroban

Thank you!

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