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Question: Let $f,g:\mathbb{R}^2 \to \mathbb{R}^2 $ be continously differentiable functions where $f\circ g$ is defined. Let $$f = (f_1,f_2)\quad\text{and}\quad g=(g_1,g_2)$$ where all $f_1,f_2,g_1,g_2:\mathbb{R}^2\to\mathbb{R}$ are functions. What is $$\frac{\partial}{\partial x_1} [f(g(x_1,x_2))]?$$

I think the derivative should be a function from $\mathbb{R}^2$ to $\mathbb{R}^2$ as well.

Let $x =(x_1,x_2).$ By multivariate chain rule, we have $$\frac{\partial}{\partial x_1} [f(g(x_1,x_2))] = \frac{\partial}{\partial x_1} [f(g_1(x),g_2(x))] = \frac{\partial f}{\partial g_1} \cdot \frac{\partial g_1}{\partial x_1} + \frac{\partial f}{\partial g_2} \cdot \frac{\partial g_2}{\partial x_1}.$$ I am not sure whether the equation above is correct.

I notice that the multiplication in the RHS of the equation above is scalar multiplication. This means that both $$\frac{\partial f}{\partial g_1}\quad\text{and}\quad \frac{\partial f}{\partial g_2}$$ is a vector in $\mathbb{R}^2.$ But I do not understand its meaning.

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Given a point $p\in{\mathbb R}^2$ the chain rule says that $$d(f\circ g)(p)=df\bigl(g(p)\bigr)\circ dg(p)\ ,$$ in Jacobian matrix terms: $$J_{f\circ g}(p)=J_f(g(p)\bigr)\cdot J_g(p)\ .$$ This implies $${\partial\over\partial x_1}(f\circ g)(p)=d(f\circ g)(p).e_1=\bigl(df\bigl(g(p)\bigr)\circ dg(p)\bigr).e_1=df\bigl(g(p)\bigr)\bigl(dg(p).e_1\bigr)\ .$$ Now $dg(p).e_1$ is the first column of the matrix $J_g(p)$; therefore we obtain $${\partial\over\partial x_1}(f\circ g)(p)=\left[\matrix{f_{1.1}&f_{1.2}\cr f_{2.1}&f_{2.2}\cr}\right]_{g(p)}\left[\matrix{g_{1.1}\cr g_{2.1}\cr}\right]_p\ ,$$ where $f_{i.k}$ means that that the component $f_i$ of $f$ is partially differentiated with respect to $x_k$.

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