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find the directional derivative of f at the point P in the indicated direction: f(xy) = xcos^2y p(2, pi/4) < 5, 1>
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i just want to make sure i have my Fx and Fy correct.. i always have difficulties with cos/sin Fx=cos^2y Fy= -2xsin^2y
\(f_x\) looks correct, but \[f_y=-2x \cos(y)\sin^2(y)\]
how come? :(
\[{\partial \over \partial y}\;\; x \cos^2(y)=2x \cos(y) \cdot \left( {\partial \over \partial y} \cos(y)\right)=-2x \cos(y) \sin(y)\]I apologize. I shouldn't have squared sin.
ohhhhh okay! cuz i have to derive the cos, that's right.. okay, cool, thank you!
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