jacobian of a dot product of 2 functions

Question

take the function $f: \mathbb R^3 \rightarrow \mathbb R^2$ an a function $g: \mathbb R^2 \rightarrow \mathbb R^2 $ defined via $$ f \begin{pmatrix} x\\ y\\ z \end{pmatrix} = \begin{pmatrix} x+z^3\\ xyz \end{pmatrix} $$ and $$ g \begin{pmatrix} s\\ t \end{pmatrix} = \begin{pmatrix} s^2 + t\\ s+t \end{pmatrix} $$ compute the vector $$ J(g \circ f) \begin{pmatrix} 1\\ 2\\ 3 \end{pmatrix}$$ , that is the jacobian of the dot product

Answer 1

I can see that you have no trouble bar silly mistakes in computing the Jacobian. More precisely, we have : $$ Df(x,y,z) = \begin{pmatrix} 1 & 0 & 3z^2 \\ zy&xz&xy \end{pmatrix} ; Dg(s,t) = \begin{pmatrix} 2s & 1 \\ 1&1 \end{pmatrix} $$

Now, the chain rule is the following : (with variables $a,b,c$ to avoid confusion) $$ D(g \circ f)(a,b,c) = Dg(f(a,b,c)) \times Df(a,b,c) $$

So, in steps , given a point $(a,b,c)$ :

• Find $f(a,b,c)$.

• Find $Df$ evaluated at $(a,b,c)$, and $Dg$ evaluated at $\color{green}{f(a,b,c)}$ , and not $\color{red}{(a,b,c)}$.

• Multiply them as matrices, in the order specified in the formula.

For example, you have $(a,b,c) = (1,2,3)$. We find that $f(1,2,3) = (28,6)$.

Now, we have $Df(1,2,3) = \begin{pmatrix} 1 & 0 & 27 \\ 6 & 3 & 2 \end{pmatrix}$.

We also have $Dg(28,6) = \begin{pmatrix} 56&1\\1&1 \end{pmatrix}$

And therefore, multiplying them in that order gives $$ \begin{pmatrix} 56&1\\1&1 \end{pmatrix} \times \begin{pmatrix} 1 & 0 & 27 \\ 6 & 3 & 2 \end{pmatrix} = \begin{pmatrix} 62 & 3 & 1514\\ 7 & 3 & 29 \end{pmatrix} $$

Which is the answer.