Question

derivadas parciais de ordem superior f(xyz) = x^2y^2cos(xz)

Answer 1

Primer orden
$f(x,y,z)=x^2y^2\cos (xz)\\ \\ f_x=2xy^2\cos(xz)-x^2y^2z\sin (xz)\\ \\ f_y=2x^2y\cos(xz)\\ \\ \\ \text{Segundo orden: }\\ \\ f_{xx}=[2y^2\cos(xz)-2x^2\sin(xz)]-[2xy^2z\sin(xz)+x^2y^2z^2\cos(xz)]\\ \\ \boxed{f_{xx}=(2y^2-x^2y^2z^2)\cos(xz)-(2x^2+2xy^2z)\sin(xz)}\\ \\ \boxed{f_{xy}=4xy\cos(xz)-2x^2yz\sin(xz)}\\ \\ \boxed{f_{yx}=4xy\cos(xz)-2x^2yz\sin(xz)}\\ \\ \boxed{f_{yy}=2x^2\cos(xz)}$