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Question Number 60426 by cesar.marval.larez@gmail.com last updated on 20/May/19

t h e f u n c t i o n i s c o n s i d e r e d

f (x, y) = e^{x y} + \frac{x}{y} + s e n ((2 x + 3 y) π) C a l c u l e :

\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial^{2} f}{\partial x^{2}}, \frac{\partial^{2} f}{\partial x \partial y} . f_{x} (0, 1), f_{y} (2, - 1), f_{x x} (0, 1), f_{x y} (2, - 1)

Commented by kaivan.ahmadi last updated on 21/May/19

\frac{\partial f}{\partial x} = {y e}^{x y} + \frac{1}{y} + 2 π c o s ((2 x + 3 y) π)

\frac{\partial f}{\partial y} = {x e}^{x y} - \frac{x}{y^{2}} + 3 π c o s (2 x + 3 y) π)

\frac{\partial^{2} f}{\partial x^{2}} = \frac{\partial}{\partial x} (\frac{\partial f}{\partial x}) = y^{2} e^{x y} - 4 π^{2} s i n (2 x + 3 y) π)

\frac{\partial^{2} f}{\partial x \partial y} = \frac{\partial}{\partial x} (\frac{\partial f}{\partial y}) = {x y e}^{x y} - \frac{1}{y^{2}} - 6 π^{2} s i n (2 x + 3 y) π)

f_{x} = \frac{\partial f}{\partial x}, \dots

f_{x} (0, 1) = e^{0} + 1 + 2 π c o s 3 π = 2 - 2 π

f_{y} (2, - 1) = 2 e^{- 2} + 3 π c o s π = \frac{2}{e^{2}} - 3 π

f_{x x} (0, 1) = e^{0} - 4 π^{2} s i n 3 π = 1

f_{x y} (2, - 1) = - 2 e^{- 2} + 1 - 6 π^{2} s i n π = \frac{- 2}{e^{2}} + 1

Commented by cesar.marval.larez@gmail.com last updated on 21/May/19

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