Find the divergence of the following vector field.
\[
F=\langle 7 y z \sin x, x z \cos y, 5 x y \cos z\rangle
\]
The divergence of $F$ is

Step 1 ：We are given the vector field \(F=\langle 7 y z \sin x, x z \cos y, 5 x y \cos z\rangle\).

Step 2 ：The divergence of a vector field \(F = \langle F1, F2, F3 \rangle\) in three dimensions is given by the formula: \(\text{div} F = \frac{\partial F1}{\partial x} + \frac{\partial F2}{\partial y} + \frac{\partial F3}{\partial z}\).

Step 3 ：Let's find the partial derivatives of each component of the vector field with respect to their respective variables.

Step 4 ：For \(F1 = 7 y z \sin x\), the partial derivative with respect to x is \(\frac{\partial F1}{\partial x} = 7 y z \cos x\).

Step 5 ：For \(F2 = x z \cos y\), the partial derivative with respect to y is \(\frac{\partial F2}{\partial y} = -x z \sin y\).

Step 6 ：For \(F3 = 5 x y \cos z\), the partial derivative with respect to z is \(\frac{\partial F3}{\partial z} = -5 x y \sin z\).

Step 7 ：Summing up these partial derivatives, we get the divergence of F: \(\text{div}_F = \frac{\partial F1}{\partial x} + \frac{\partial F2}{\partial y} + \frac{\partial F3}{\partial z} = 7 y z \cos x - x z \sin y - 5 x y \sin z\).

Step 8 ：Final Answer: The divergence of the vector field \(F=\langle 7 y z \sin x, x z \cos y, 5 x y \cos z\rangle\) is \(\boxed{-5 x y \sin z - x z \sin y + 7 y z \cos x}\).