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Alright, I'm trying to fine the gradient of the function and the maximum value of the directional derivative.

$z = \frac{y}{(x^2 + y^2 )}\text{ at }(1,1)$

So far, I've gotten the following values for partial derivates, but I can't guarantee they are correct:

$F_x = \frac{-2xy}{(x^2 + y^2 )^2}$ $F_y = \frac{-2y^2}{( x^2 + y^2 )^2} + \frac1{x^2 + y^2}$

GradientF(x,y) = \frac{-2xy}{( x^2 + y^2 )^2 + (-2y^2)/( x^2 + y^2 )^2} + 1/( x^2 + y^2 )^2}$$ GradientF(1,1) = -2/4i + (-2/4)j + (1/2)k = <-2/4, -2/4, 1/2> ||F(1,1)|| = <-2/4, -2/4, 1/2> (1/sqrt(3/4))

I get <-2/3, -2/3, 2/3> but this is wrong...

I should be getting <1/2, 0> at max value of 1/2 according to my friend who ceases to explain anything to me.

Any help? Thanks.

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    I actually mis-wrote that. I edited it now, it wasn't squared on the bottom. Sorry!2011-05-11

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To evaluate partial derivatives of a function $f(x,y)$, you fix one of the variables as the constant, and derivate with respect to the other variable. For instance, if you're trying to find, say, $\partial_x f$, treat $y$ as a constant; you can temporarily change $y$ it for $a$ if you wish so as to see it more clearly.

The maximum value of the directional derivative will occur in the direction along the gradient vector (at a given point). This maximum value will be the norm of the gradient vector (at that point) -- just review the definition of directional derivative, it's a dot product between the gradient vector and a unit vector that gives the "direction".

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    Do the dot product of the vector with itself, then take the square root.2011-05-11