Answer
Vector along the gradient: $\mathbf{i}-\mathbf{j}-\mathbf{k}=\mathbf{V}$
Unit vecotr $\mathbf{v}=\frac{1}{\sqrt{3}}(\mathbf{i}-\mathbf{j}-\mathbf{k})$
Rate of change: $\|\nabla f(4,2,2)\|=\frac{\sqrt{3}}{8}$
Work Step by Step
$f$ increases more quickly in the positive direction of $\nabla f(x, y, z)$
\[
\begin{array}{l}
\frac{y+z}{x^{2}+(y+z)^{2}}=f_{x}(x, y, z) \\
\frac{1}{8}=f_{x}(4,2,2) \\
-\frac{x}{x^{2}+(y+z)^{2}}=f_{y}(x, y, z) \\
-\frac{1}{8} =f_{y}(4,2,2) \\
f_{z}(x, y, z)=-\frac{x}{x^{2}+(y+z)^{2}} \\
f_{z}(4,2,2)=-\frac{1}{8}
\end{array}
\]
$\frac{1}{8}(\mathbf{i}-\mathbf{j}-\mathbf{k})=\nabla f(4,2,2)$
Vector along the gradient: $\mathbf{i}-\mathbf{j}-\mathbf{k}=\mathbf{V}$
Unit vecotr $\mathbf{v}=\frac{1}{\sqrt{3}}(\mathbf{i}-\mathbf{j}-\mathbf{k})$
Rate of change: $\|\nabla f(4,2,2)\|=\frac{\sqrt{3}}{8}$
