91Ó°ÊÓ

We are given \(F(x, y, z)=e^{xyz}\). To write it as a composite function, we can let \(g(x, y, z) = xyz\). Now we need to find a function \(f(u)\) such that \(F(x,y,z) = f \circ g = f(g(x,y,z))\). In our case, we want \(f(u) = e^u\). So, $$ F(x, y, z) = f \circ g = f(g(x, y, z)) = f(xyz) = e^{xyz} $$ Thus, the composite functions are \(f(u) = e^u\) and \(g(x, y, z) = xyz\).

To relate the gradient of F to the gradient of g, we first find the gradient of F and the gradient of g. The gradient of F is given by: $$ ∇F = \begin{bmatrix} \frac{\partial F}{\partial x}\\ \frac{\partial F}{\partial y}\\ \frac{\partial F}{\partial z} \end{bmatrix} $$ Since \(F(x, y, z) = e^{xyz}\), we compute the partial derivatives: $$ \frac{\partial F}{\partial x} = ye^{xyz}, \frac{\partial F}{\partial y} = xe^{xyz}, \frac{\partial F}{\partial z} = xye^{xyz} $$ Thus, the gradient of F is: $$ ∇F = \begin{bmatrix} ye^{xyz}\\ xe^{xyz}\\ xye^{xyz} \end{bmatrix} $$ Now, let's find the gradient of g: $$ ∇g = \begin{bmatrix} \frac{\partial g}{\partial x}\\ \frac{\partial g}{\partial y}\\ \frac{\partial g}{\partial z} \end{bmatrix} $$ Since \(g(x, y, z) = xyz\), we compute the partial derivatives: $$ \frac{\partial g}{\partial x} = yz, \frac{\partial g}{\partial y} = xz, \frac{\partial g}{\partial z} = xy $$ Thus, the gradient of g is: $$ ∇g = \begin{bmatrix} yz\\ xz\\ xy \end{bmatrix} $$ To relate the gradient of F to the gradient of g, we can observe that the gradient of F can be written as a product of the gradient of g and F(x,y,z). That is: $$ ∇F = F(x,y,z) \cdot ∇g $$ Therefore, we find that the gradient of F can be related to the gradient of g by multiplying the elements of the gradient of g by the value of the function F(x, y, z).