First, take the first order partial derivatives:
f_x = sin(x+y) + x cos(x+y), and f_y = x cos(x+y).
For the critical points, set these equal to 0:
sin(x+y) + x cos(x+y) = 0, and x cos(x+y) = 0.
Hence, sin(x+y) = 0 and x cos(x+y) = 0.
(i) If x = 0, then we need sin(0+y) = 0 ==> y = nπ for any integer n.
(ii) Otherwise, we need sin(x+y) = 0 and cos(x+y) = 0, which has no solution.
So, the critical points are at (x, y) = (0, nπ) for any integer n.
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Next, we classify these with the Second Derivative Test.
Since f_x = sin(x+y) + x cos(x+y), and f_y = x cos(x+y), we have
f_xx = 2 cos(x+y) - x sin(x+y)
f_yy = -x sin(x+y)
f_xy = cos(x+y) - x sin(x+y)
==> D = f_xx * f_yy - (f_xy)^2
..........= [2 cos(x+y) - x sin(x+y)][cos(x+y) - x sin(x+y)] - (-x sin(x+y)^2).
Note that D(0, nπ) = 2 cos^2(nπ) = 2 > 0, and f_xx (0, nπ) = 2 cos(nπ).
(i) If n is even, then D(0, nπ) > 0, and f_xx (0, nπ) = 2 > 0.
==> We have a local minimum.
(ii) If n is odd, then D(0, nπ) > 0, and f_xx (0, nπ) = -2 < 0.
==> We have a local maximum.
I hope this helps!
f_x = sin(x+y) + x cos(x+y), and f_y = x cos(x+y).
For the critical points, set these equal to 0:
sin(x+y) + x cos(x+y) = 0, and x cos(x+y) = 0.
Hence, sin(x+y) = 0 and x cos(x+y) = 0.
(i) If x = 0, then we need sin(0+y) = 0 ==> y = nπ for any integer n.
(ii) Otherwise, we need sin(x+y) = 0 and cos(x+y) = 0, which has no solution.
So, the critical points are at (x, y) = (0, nπ) for any integer n.
------------------
Next, we classify these with the Second Derivative Test.
Since f_x = sin(x+y) + x cos(x+y), and f_y = x cos(x+y), we have
f_xx = 2 cos(x+y) - x sin(x+y)
f_yy = -x sin(x+y)
f_xy = cos(x+y) - x sin(x+y)
==> D = f_xx * f_yy - (f_xy)^2
..........= [2 cos(x+y) - x sin(x+y)][cos(x+y) - x sin(x+y)] - (-x sin(x+y)^2).
Note that D(0, nπ) = 2 cos^2(nπ) = 2 > 0, and f_xx (0, nπ) = 2 cos(nπ).
(i) If n is even, then D(0, nπ) > 0, and f_xx (0, nπ) = 2 > 0.
==> We have a local minimum.
(ii) If n is odd, then D(0, nπ) > 0, and f_xx (0, nπ) = -2 < 0.
==> We have a local maximum.
I hope this helps!