# Easy question on the partial derivative of a compound function

Let $F : \mathbb{R}^2 \rightarrow \mathbb{R}^2, u : \mathbb{R}^2 \rightarrow \mathbb{R},$ both $C^{\infty}$ and $g (x,y) = u(F(x,y))$. What is $\partial_x g$? And $\partial^2_{x,x} g$?

Let $F(x,y) = (F_1(x,y), F_2(x,y)), u=u(a,b)$.
$$DF = \begin{pmatrix} \partial_x F_1 & \partial_y F_1\\ \partial_x F_2 & \partial_y F_2 \end{pmatrix}, \; Du= \begin{pmatrix} \partial_a u & \partial_b u \end{pmatrix}$$ How should I apply $((Du) \circ F) \cdot (DF)$ ?

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You presume you do know the multi-varate chain rule, i.e., $D(u \circ F) = ((Du) \circ F) \cdot (DF)$, where $D$ is the differential operator? –  Johannes Kloos Jun 10 '12 at 13:03

Try explicitly

$$\partial_x g = \frac{ \partial u }{\partial F_1}\frac{\partial F_1}{\partial x} + \frac{ \partial u }{\partial F_2}\frac{\partial F_2}{\partial x}$$ where $$\frac{\partial u}{\partial F_1} = \frac{ \partial u}{\partial a} \quad \mbox{ and } \frac{\partial u}{\partial F_2} = \frac{ \partial u}{\partial b}$$ where $u=u(a,b)$. Similarly for the $y$.

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