In my vector calculus class, we are studying multivariate vector-valued functions and I have come across this exercise
Let $f: \mathbb{R}^2 \to \mathbb{R}^3 $, $g: \mathbb{R}^3 \to \mathbb{R}^2$, and $u: \mathbb{R}^3 \to \mathbb{R}^3$ be defined as follows $$ f(y) = \begin{pmatrix} y_1 - e^{y_2} \\ y_2 \\ 2(y_1-2)^3 \end{pmatrix} $$ and $$ g(x) = \begin{pmatrix} x_1+2x_2+3x_3 \\ \ln(x_1+2x_2) \end{pmatrix} $$ and $$ u(x) = f(g(x)) $$ and let $x_0 = \begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix}$ and $y_0 = g(x_0) = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$. We are asked to find an explicit formula for $u(x)$ in terms of $x$, and to compute the Jacobians $Df(y_0),Dg(x_0),Du(x_0)$ and to verify that $$ Du(x_0) = Df(g(x_0)) \cdot Dg(x_0) $$ as the chain rule.
I am a complete novice to multivariate high-dimensional functions and to Jacobian matrices which I am unsure how to compute and cannot understand the last identity given to verify (the chain rule). I would appreciate help with this problem. I thank all helpers.
Given two positive integers, $n$ and $m$, and a function $h : \mathbb R^n \to \mathbb R^m$ given by $$h(u) = \begin{pmatrix} h_1(u) \\ h_2(u) \\ \vdots \\ h_m(u) \end{pmatrix} \textrm{ where } u = \begin{pmatrix} u_1 \\ \vdots \\ u_n \end{pmatrix},$$ the derivative of $h$ at $a \in \mathbb R^n$, denoted by $Dh(a)$, is the $m \times n$ matrix whose $i$-th row is $$\nabla h_i(a) = \Big( \frac{\partial h_i}{\partial u_1}\Big|_{u=a} \ \ \ \frac{\partial h_i}{\partial u_2}\Big|_{u=a} \ \ \ \cdots \ \ \ \frac{\partial h_i}{\partial u_n}\Big|_{u=a} \Big).$$ Does this help you?