Exercise 1

\[f(x, y) = x^2 + y^2 - \sin (xy)\] \[\frac{\partial f(x,y)}{\partial x} = 2x - y \cos (xy)\] \[\frac{\partial f(x,y)}{\partial y} = 2y - x \cos (xy)\] \[\frac{\partial^2 f(x,y)}{\partial x^2} = 2 + y^2 \sin (xy)\] \[\frac{\partial^2 f(x,y)}{\partial y^2} = 2 + x^2 \sin (xy)\] If \(f(x, y)\) is partially differentiable and if the derivatives are continous, then the sequence of the partial derivatives can be changed (Theorem of Schwarz): \[\frac{\partial^2 f(x,y)}{\partial x\, \partial y} = \frac{\partial^2 f(x,y)}{\partial y \, \partial x} = - \cos (xy) + x y \sin (xy)\]

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\[f(x,y ) = \frac{x}{y} \mathrm{e}^{(x^2+y^2)} \] \[\frac{\partial f(x,y)}{\partial x} = \frac{1}{y} \mathrm{e}^{(x^2+y^2)} + \frac{x}{y} \mathrm{e}^{(x^2+y^2)} 2x\] \[= \frac{1+2x^2}{y} \mathrm{e}^{(x^2+y^2)} \] \[\frac{\partial f(x,y)}{\partial y} = - \frac{x}{y^2} \mathrm{e}^{(x^2+y^2)} + \frac{x}{y}\mathrm{e}^{(x^2+y^2)} 2y\] \[= x \mathrm{e}^{(x^2+y^2)}(2-\frac{1}{y^2})\] \[\frac{\partial^2 f(x,y)}{\partial x^2} = \frac{4x}{y} \mathrm{e}^{(x^2+y^2)} + \frac{2x^2+1}{y}\mathrm{e}^{(x^2+y^2)} 2x\] \[=\frac{2x}{y} \mathrm{e}^{(x^2+y^2)}(2x^2 +3)\] \[\frac{\partial^2 f(x,y)}{\partial y^2} = x \left( \mathrm{e}^{(x^2+y^2)} 2 y (1 - \frac{1}{y^2}) + \mathrm{e}^{(x^2+y^2)} \frac{2}{y^3} \right)\] \[= 2x \mathrm{e}^{(x^2+y^2)} \left( \frac{1}{y^3} - \frac{1}{y} +2y \right)\] \[\frac{\partial^2 f(x,y)}{\partial x \partial y} = \frac{\partial}{\partial x} \left( x\mathrm{e}^{(x^2+y^2)}(2-\frac{1}{y^2}) \right)\] \[=\left( \mathrm{e}^{(x^2+y^2)} + x \mathrm{e}^{(x^2+y^2)} 2x \right) (2-\frac{1}{y^2})\] \[=\mathrm{e}^{(x^2+y^2)} (1+2x^2)(2-\frac{1}{y^2})\]

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\[f(x, y) = \mathrm{e}^x \cos y\] \[\frac{\partial f}{\partial x}=\mathrm{e}^x \cos y\] \[\frac{\partial^2 f}{\partial x^2}=\mathrm{e}^x \cos y\] \[\frac{\partial f}{\partial y}=-\mathrm{e}^x \sin y\] \[\frac{\partial^2 f}{\partial y^2}=-\mathrm{e}^x \cos y\] \[\frac{\partial^2 f}{\partial y \partial x}=-\mathrm{e}^x \sin y\]