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I'm having trouble with a math problem. I'm supposed to find the directional derivative for $f(x,y) = x^2cos(y)$ at $(x,y) = \big(0, \frac{\pi}{2}\big)$ in the direction $w = (1,-1)$ and also find the direction vector for which directional derivative is maximal.

What I've done so far is this:

$f_x(x,y) = 2x\cos{y}$

$f_y(x,y) = -x^2\sin{y}$

$|w| = \sqrt{1^2+(-1)^2}=\sqrt{2}$

$u=\frac{w}{|w|}=(1/\sqrt{2}, -1/\sqrt{2})$

So $D_u\,f(x,y) = \nabla f(x,y) \cdot u$

$\nabla f(x,y) = (2x\cos{y}, -x^2\sin{y})$

$\nabla f(0,\pi/2) = (2(0)\cos{\frac{\pi}{2}}, -(0)^2\sin{\frac{\pi}{2}} = (0, 0)$
$D_u\,f(x,y) = (0,0) \cdot (1/\sqrt{2}, -1/\sqrt{2}) = 0$

So the directional derivative is 0, meaning the function does not increase or decrease. Am I doing this correctly? How am I supposed to do the second part if the first part is 0? Thanks!

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Yes, your answer is right.

For the second part, you can just say that the direction vector for which directional derivative is maximal does not exists.

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