# Maximizing directional derivatives?

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.

-
Your partial with respect to $x$ is missing a factor of $\frac2{x^2+y^2}$... – J. M. May 11 '11 at 3:57
How? I just did it again and got the same answer? – jtm22 May 11 '11 at 4:01
Maybe you can post your solution in detail? Then we can debug... – J. M. May 11 '11 at 4:03
OK, mind editing in my mistakes that I make when formatting? Sorry, I'm new here =( – jtm22 May 11 '11 at 4:05
Edited in. Now for formatting – jtm22 May 11 '11 at 4:09

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.
@jtm22: Where's that $k$ coming from? Don't you have two variables only? The gradient should be a two-dimensional vector. You just have to calculate its norm. – Weltschmerz May 11 '11 at 4:21