Multivariable chain rule and Jacobian

Question

Let f : $R^3$ → $R^2$ be given by f(r, s, t) = ($r^3$s + $t^2$, rst). Let g : $R^2$ → $R^{11}$ and h : $R^{11}$ → $R^4$ be two differentiable functions. Compute $J_{h◦g◦f}$ (0, 21, 0).

I've tried the question and now I'm stuck, this is what I have done so far.

Current workings :

$J_{h◦g◦f}$ = $J_{h}(g(f(r,s,t)))$ ⋅ $J_{g}(f(r,s,t))$ ⋅ $J_{f}(r,s,t)$

f(r,s,t) = $\begin{bmatrix}f_1(r,s,t)\\f_2(r,s,t)\end{bmatrix}$ = $\begin{bmatrix}r^3s + t^2\\rst\end{bmatrix}$

$J_f$ = $\begin{bmatrix}3r^2s & r^3 & 2t\\st & rt & rs\end{bmatrix}$

I understand that my next step should be finding the Jacobians of $J_g$ and $J_h$. Using these individual matrices, I can multiply them and find $J_{g◦f}$ and $J_{h◦g◦f}$.

But I'm not sure how to find $J_g$ and $J_h$ because I don't know how to express them in terms of r,s,t.

Any help please?

Answer 1

Note that $J_f= \begin{pmatrix} 3r^2s & r^3 & 2t\\st & rt & rs \end{pmatrix}$ evaluated at $(0,21,0)$ is the $(2\times 3)$ $0$-matrix.

We conclude, that $J_{h\circ g\circ f}$ evaluated at $(0,21,0)$ is

\begin{align*} J_{h\circ g\circ f}(0,21,0) &=J_{h}(g(f(0,21,0)))\cdot J_{g}(f(0,21,0)) \cdot J_{f}(0,21,0)\\ &=J_{h}(g(f(0,21,0)))\cdot J_{g}(f(0,21,0))\cdot \begin{pmatrix} 0& 0& 0\\0& 0& 0 \end{pmatrix}\\ &=(0)_{{1\leq j\leq 4}\atop{1\leq k\leq 11}} \end{align*}

the $(4\times 11)$ $0$-matrix.