Let $h: \mathbb{R}^3 \rightarrow \mathbb{R}$ such that $h(x,y,z)=g(x^2-y^2,y^2-z^2)$ and $g: \mathbb{R}^2 \rightarrow \mathbb{R}$ a differentiable function such that $\nabla g(0,0)=(1,2)$. Determine the directional derivative of $h$ in the point $(1,1,1)$ along $\vec{v}=(2,0,1)$.
Though it might be possible to solve this problem without the Jacobian Matrix, I would really like to know how to aproach this problem with it. I'm having difficulty, since the dimensions of the matrix don't correspond to what's expected.
Best Answer
The "trick" here is that you are really working with the composition $g\circ f$, where $g$ is as defined in the problem and $f:\mathbb{R}^3\to\mathbb{R}^2$ is given by $f(x,y,z)=(x^2-y^2,y^2-z^2)$. Working with these you should get a 1x2 matrix (the Jacobian of $g$, evaluated at $f(1,1,1)=(0,0)$) times a 2x3 matrix (the Jacobian of $f$, evaluated at $(1,1,1)$):
$$ \begin{bmatrix} 1 & 2 \end{bmatrix}\begin{bmatrix} 2x & -2y & 0 \\ 0 & 2y & -2z \end{bmatrix}_{(1,1,1)} = \begin{bmatrix}2x & 2y & -4z\end{bmatrix}_{(1,1,1)}=\begin{bmatrix}2 & 2 & -4\end{bmatrix} $$
And the dimensions of this product are as expected since the composition $g\circ f$ maps from $\mathbb{R}^3$ to $\mathbb{R}$. You can then treat the resulting 1x3 matrix as a gradient to compute the directional derivative:
$$ \mathbf{D}_{\overrightarrow{v}}(g\circ f)(1,1,1) = \frac{1}{\sqrt{5}} \begin{bmatrix}2 & 2 & -4\end{bmatrix}\begin{bmatrix}2 \\ 0 \\ 1\end{bmatrix}=0 $$