Get the answer to "Find all the second partial derivatives. f(x,y)=sin2(mx+ny)"

Jason Farmer

Answered question

2021-06-08

Find all the second partial derivatives.

f (x, y) = s i n^{2} (m x + n y)

dieseisB

Skilled2021-06-09Added 85 answers

The answer is given below:

Jeffrey Jordon

Expert2021-09-09Added 2605 answers

$f = [\sin (m x + n y)]^{2}$

$f_{x} = 2 m \sin (m x + n y) \cdot \cos (m x + n y)$

$f_{x x} = 2 m^{2} \cos (m x + n y) \cdot \cos (m x - n y) - 2 m^{2} \sin (m x + n y) \sin (m x + n y)$

$f_{x x} = 2 m = \cos^{2} (m x + n y) - 2 m^{2} \sin^{2} (m x + n y)$

$f_{x y} = 2 m n \cos (m x + n y) \cos (m x + n y) - 2 m n \sin (m x + n y) \sin (m x + n y)$

$f_{x y} = 2 m n \cos^{2} (m x + n y) - 2 m n \sin^{2} (m x + n y)$

$f_{y} = 2 n \sin (m x + n y) \cos (m x + n y)$

$f_{y y} = 2 n^{2} \cos (m x + n y) \cos (m x + n y) - 2 n^{2} \sin (m x + n y) \sin (m x + n y)$

$f_{y y} = 2 n^{2} \cos^{2} (m x + n y) - 2 n^{2} \sin^{2} (m x + n y)$

$f_{y x} = 2 m n \cos (m x + n y) \cos (m x + n y) - 2 m n \sin (m x + n y)$

$f_{y x} = 2 m n \cos^{2} (m x + n y) - 2 m n \sin^{2} (m x + n y)$

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