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Set $f(x,y,z) = xyz$ and $g(s,t) = (3s+st,s,t)$. Then define F(s,t) as the composition $(f\circ g)(s,t)$. Determine the derivative of $F$ at the origin.

My book defines the derivative $Df(a)$ of $f:\mathbb{R}^n\to\mathbb{R}^m$ at $a$ as equal to the jacobi matrix, rather $$Df(a) = \begin{bmatrix} \frac{\partial f_i}{\partial x_j} \end{bmatrix}\bigg|_{x=a}$$ where $f_i$ is the $i$th coordinate funciton.

I feel like I should first should that $f$ and $g$ are differentiable, which will then give the composition is differentiable (from there I'll use chain rule). So starting with differentiability, I know that a function $f$ in $\mathbb{R}^n$ is differentiable if all of its partial derivatives both exist and are continuous. Thus, I can use the definition of partial derivative for a function $f$ at $a$: $$\lim_{t\to0}\frac{f(a+te_i)-f(a)}{t}$$ for the $i$th unit basis vector $e_i$, to find the partial derivatives of the functions defined above, $f$ and $g$, at the origin.

Now what I am stuck on showing that each partial derivative of the functions $f(x,y,z) = xyz$ and $g(s,t) = (3s+st,s,t)$ are continuous at the origin. I believe that I am expected to give an $\epsilon$-$\delta$ type proof, or at least as rigorous. Could I have some help on this please?

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    $\begingroup$ That seems like way too much work to me. Both functions are quite obviously differentiable everywhere, so you need but multiply the two Jacobian matrices, taking care to evaluate $Df$ at $g(0,0)$. You can save yourself a bit of working by noting that neither determinant has any linear terms, so you’re going to end up with a matrix of zeros. $\endgroup$ – amd Nov 7 '17 at 23:44
  • $\begingroup$ Yeah it seems like quite a bit of work, but I feel that saying it's obvious won't necessarily do the job. $\endgroup$ – user500472 Nov 7 '17 at 23:47
  • $\begingroup$ Each individual component function is a sum, product &c of very obviously everywhere-differentiable functions. That’s what makes differentiability “obvious.” $\endgroup$ – amd Nov 7 '17 at 23:51
  • $\begingroup$ $g(0,0)=(0,0,0)$ so $f\circ g(0,0)=0$. so $Df(g(\vec 0))\circ Dg(\vec 0)=0$ because the matrix of parcial derivatives of $f$ is zero at the orgin. $\endgroup$ – Matematleta Nov 7 '17 at 23:52
  • $\begingroup$ @ChilangoIncomprendido Except that $D(f\circ g)$ is not a scalar. $\endgroup$ – amd Nov 7 '17 at 23:57

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