I am aware this could be a dumb question but I got myself stuck in some complicated calculus that I have greatly simplified to focus on my lack of understanding.

My goal is to compute the derivative of the following composition : \begin{align} (f\circ g)'(0) \end{align} Which is the composition evaluated at 0. $f: \mathbb{R^n}\rightarrow \mathbb{R}$ and $g: \mathbb{R}\rightarrow \mathbb{R^n}$ so the total function $f\circ g: \mathbb{R}\rightarrow \mathbb{R}$

Intuitively using the chain rule : \begin{align} (f\circ g)' = (f'\circ g)g' \end{align}

So in my particular case since $f$ is from $\mathbb{R^n}$ to derive it I need to take the sum of the partials multiplied by the derivative of the corresponding entry. For g I just need to derive each entry. The problem is that the derivative of g lead to n values that I can't evaluate on 0.

Can someone provide me a way to solve this ?


$g'$ has $n$ components, and $f$ has $n$ partial derivatives. You need to multiply each component of $g'$ with the corresponding partial derivative of $f$ (evaluated at $g(0)$) and then take the sum.

In general, the derivative of a function $\mathbb R^m\to\mathbb R^n$ is an $m\times n$ matrix, and the chain rule is a matrix multiplication.

  • $\begingroup$ In particular, $\mathrm df$ is a row vector and $\mathrm dg$ is a column vector here. $\endgroup$ – amd Nov 27 '15 at 22:33
  • $\begingroup$ I understand thank you ! So this has nothing to do with the total derivative of $f$ then ? On the wikipedia page of the chain rule, they say that the generalization in higher dimension involve the total derivative. $\endgroup$ – user149705 Nov 27 '15 at 22:41
  • $\begingroup$ As far as I understand total derivative it's a special case of the multidimensional chain rule. $\endgroup$ – Justpassingby Nov 28 '15 at 9:49

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