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

$$thefunctionisconsidered$$$$f(x,y)=e^{xy}+\frac{x}{y}+sen\left((2x+3y)\pi \right)Calcule:$$$$\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{{\partial }^{2}f}{\partial x^2},\frac{{\partial }^{2}f}{\partial x\partial y}.f_x(0,1),f_y(2,-1),f_{xx}(0,1),f_{xy}(2,-1)$$

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

$$\frac{\partial f}{\partial x}={ye}^{xy}+\frac{1}{y}+2\pi cos\left((2x+3y)\pi \right)$$$$\frac{\partial f}{\partial y}={xe}^{xy}-\frac{x}{y^2}+3\pi cos(2x+3y)\pi )$$$$\frac{{\partial }^{2}f}{\partial x^2}=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial x}\right)=y^2{e}^{xy}-4{\pi }^{2}sin(2x+3y)\pi )$$$$\frac{{\partial }^{2}f}{\partial x\partial y}=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial y}\right)={xye}^{xy}-\frac{1}{y^2}-6{\pi }^{2}sin(2x+3y)\pi )$$$$f_x=\frac{\partial f}{\partial x},...$$$$f_x(0,1)=e^0+1+2\pi cos3\pi =2-2\pi$$$$f_y(2,-1)=2e^{-2}+3\pi cos\pi =\frac{2}{e^2}-3\pi$$$$f_{xx}(0,1)=e^0-4{\pi }^{2}sin3\pi =1$$$$f_{xy}(2,-1)=-2e^{-2}+1-6{\pi }^{2}sin\pi =\frac{-2}{e^2}+1$$

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

$$thankssirforthesolution$$

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