need my work checked plz :)

**mathlovet** · October 6th 2008, 09:35 PM

u= ln(sqrt(x^2+y^2))

u(x) = x / (2sqrt((x^2+y^2))
u(xx) = (1-2x^2)/(2(sqrt(x^2+Y^2))^3)

u(y)= y / (2sqrt((x^2+y^2))
u(yy)= (1-2y^2)/(2(sqrt(x^2+Y^2))^3)

i havent done derivatives for a while now

.. not sure it is correct..

**lllll** · October 6th 2008, 10:11 PM

you could use some of the properties of logarithms for this one:

$u = \ln(\sqrt{x^2+y^2}) = \frac{1}{2}\ln(x^2+y^2)$

so then:

$u(x) = \frac{1}{2} \times \frac{1}{x^2+y^2} \times 2x = \frac{x}{x^2+y^2}$

$u(xx) = \frac{(1)(x^2+y^2) - 2x(x)}{(x^2+y^2)^2} = \frac{-x^2+y^2}{(x^2+y^2)^2}$

$u(y) = \frac{1}{2} \times \frac{1}{x^2+y^2} \times 2y = \frac{y}{x^2+y^2}$

$u(yy) = \frac{(1)(x^2+y^2) - 2y(y)}{(x^2+y^2)^2} = \frac{x^2-y^2}{(x^2+y^2)^2}$

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