derivadas parciais de ordem superior f(xyz) = x^2y^2cos(xz)
daanigc:
segunda ordem
Soluções para a tarefa
Respondido por
1
Primer orden
![f(x,y,z)=x^2y^2\cos (xz)\\ \\
f_x=2xy^2\cos(xz)-x^2y^2z\sin (xz)\\ \\
f_y=2x^2y\cos(xz)\\ \\ \\
\text{Segundo orden: }\\ \\
f_{xx}=[2y^2\cos(xz)-2x^2\sin(xz)]-[2xy^2z\sin(xz)+x^2y^2z^2\cos(xz)]\\ \\
\boxed{f_{xx}=(2y^2-x^2y^2z^2)\cos(xz)-(2x^2+2xy^2z)\sin(xz)}\\ \\
\boxed{f_{xy}=4xy\cos(xz)-2x^2yz\sin(xz)}\\ \\
\boxed{f_{yx}=4xy\cos(xz)-2x^2yz\sin(xz)}\\ \\
\boxed{f_{yy}=2x^2\cos(xz)}](https://tex.z-dn.net/?f=f%28x%2Cy%2Cz%29%3Dx%5E2y%5E2%5Ccos+%28xz%29%5C%5C+%5C%5C%0Af_x%3D2xy%5E2%5Ccos%28xz%29-x%5E2y%5E2z%5Csin+%28xz%29%5C%5C+%5C%5C%0Af_y%3D2x%5E2y%5Ccos%28xz%29%5C%5C+%5C%5C+%5C%5C%0A%5Ctext%7BSegundo+orden%3A+%7D%5C%5C+%5C%5C%0Af_%7Bxx%7D%3D%5B2y%5E2%5Ccos%28xz%29-2x%5E2%5Csin%28xz%29%5D-%5B2xy%5E2z%5Csin%28xz%29%2Bx%5E2y%5E2z%5E2%5Ccos%28xz%29%5D%5C%5C+%5C%5C%0A%5Cboxed%7Bf_%7Bxx%7D%3D%282y%5E2-x%5E2y%5E2z%5E2%29%5Ccos%28xz%29-%282x%5E2%2B2xy%5E2z%29%5Csin%28xz%29%7D%5C%5C+%5C%5C%0A%5Cboxed%7Bf_%7Bxy%7D%3D4xy%5Ccos%28xz%29-2x%5E2yz%5Csin%28xz%29%7D%5C%5C+%5C%5C%0A%5Cboxed%7Bf_%7Byx%7D%3D4xy%5Ccos%28xz%29-2x%5E2yz%5Csin%28xz%29%7D%5C%5C+%5C%5C%0A%5Cboxed%7Bf_%7Byy%7D%3D2x%5E2%5Ccos%28xz%29%7D%0A%0A "f(x,y,z)=x^2y^2\cos (xz)\\ \\
f_x=2xy^2\cos(xz)-x^2y^2z\sin (xz)\\ \\
f_y=2x^2y\cos(xz)\\ \\ \\
\text{Segundo orden: }\\ \\
f_{xx}=[2y^2\cos(xz)-2x^2\sin(xz)]-[2xy^2z\sin(xz)+x^2y^2z^2\cos(xz)]\\ \\
\boxed{f_{xx}=(2y^2-x^2y^2z^2)\cos(xz)-(2x^2+2xy^2z)\sin(xz)}\\ \\
\boxed{f_{xy}=4xy\cos(xz)-2x^2yz\sin(xz)}\\ \\
\boxed{f_{yx}=4xy\cos(xz)-2x^2yz\sin(xz)}\\ \\
\boxed{f_{yy}=2x^2\cos(xz)}")
Perguntas interessantes