f(x,y)=ex+ysinx+9x2+2yf (x,y)=e^x+y sinx +9x^2+2yf(x,y)=ex+ysinx+9x2+2y
fxy=(fx)y=∂∂y(∂f∂x)f_{xy}=(f_x)_y= \frac{\partial }{\partial y}(\frac{\partial f}{\partial x})fxy=(fx)y=∂y∂(∂x∂f)
=∂∂y(ex+ycosx+18x)=\frac{\partial }{\partial y}(e^x+ycosx+18x)=∂y∂(ex+ycosx+18x)
=cosx=cosx=cosx
Thus, fxy(1,2)=cos(1)f_{xy}(1,2)=\cos(1)fxy(1,2)=cos(1)
fyx=(fy)x=∂∂x(∂f∂y)f_{yx}=(f_y)_x= \frac{\partial }{\partial x}(\frac{\partial f}{\partial y})fyx=(fy)x=∂x∂(∂y∂f)
=∂∂x(sinx+2)=\frac{\partial }{\partial x}(sinx+2)=∂x∂(sinx+2)
Thus, fyx(1,2)=cos(1)f_{yx}(1,2)=cos(1)fyx(1,2)=cos(1)
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