Let $U \subset \mathbb{R}^n$ be open, and let $\gamma: \mathbb{R} \to U$ and $f: U \to \mathbb{R}$ be to functions that are differentiable at least twice.

I want to show that $\frac{d^2}{dt^2}(f \circ \gamma)(t)$ is then given by:

$$\frac{d^2}{dt^2}(f \circ \gamma)(t) = \dot{\gamma}(t)^t H f(\gamma(t))\dot{\gamma}(t)+ d f(\gamma(t))\dot{\dot{\gamma}}(t)$$

(where $H f(x)$ is the Hesse matrix of $f$ at $x \in U$).

I must admit that I don't really know how to get started. It looks like one could utilise the multivariable chain rule?


$\gamma(t+h) = \gamma(t)+ \gamma'(t)h+\frac{1}{2}\gamma''(t)h^2+o(h^2)$

$f(\vec{t} + \vec{h}) = f(\vec{t}) + df\big|_{\vec{t}}(\vec{h}) + \frac{1}{2}\vec{h}^\top H(\vec{t}) \vec{h} + o(|\vec{h}|^2) $

These are both by the multivariable taylor's series.

Composing these two should give you the beginning of the taylor series for $(f \circ \gamma)(t)$, with a rigorous bound on the error. This should justify your formula.

Now just compose these approximations:

$\begin{align*} f(\gamma(t+h)) &\approx f\left( \gamma(t) + \gamma'(t)h + \frac{1}{2} \gamma''(t)h^2\right)\\ &\approx f(\gamma(t))+df\big|_{\gamma(t)}\left(\gamma'(t)h + \frac{1}{2} \gamma''(t)h^2 \right) \\ &\hphantom{hhhhh}+ \left(\gamma'(t)h + \frac{1}{2} \gamma''(t)h^2\right)^\top Hf\big|_{\gamma(t)}\left(\gamma'(t)h + \frac{1}{2} \gamma''(t)h^2 \right)\\ &=f(\gamma(t))+df\big|_{\gamma(t)}(\gamma'(t))h + \frac{1}{2}\left( df\big|_{\gamma(t)}\left(y''(t)\right) + \gamma'(t)^\top H(f)\big|_{\gamma(t)}(\gamma'(t)) \right)h^2 + \textrm{higher order terms} \end{align*} $

So the second derivative is $df\big|_{\gamma(t)}\left(y''(t)\right) + \gamma'(t)^\top H(f)\big|_{\gamma(t)}(\gamma'(t))$ as your thought.

I didn't track the little o terms because it got too messy. I hope you can see that we are just composing taylor series. A good exercise would be to find $(f \circ g)''(x)$ for $f,g :\mathbb{R} \to \mathbb{R}$ this way as well.

  • $\begingroup$ Was my suggestion helpful? Do you need more details? $\endgroup$ Jul 1 '15 at 10:54
  • $\begingroup$ Well; thanks but I actually have trouble finishing it so far. I think I understand the first part, but I'm not so used to working with Taylor series with multiple variables, so some more details on the second part part and why this already justifies the formula would be appreciated. :) $\endgroup$
    – moran
    Jul 1 '15 at 14:07
  • $\begingroup$ @moran I updated my answer. $\endgroup$ Jul 2 '15 at 15:41

For the first derivation we get:

$d/dt f(\gamma(t)) = <\nabla f(\gamma_1(t),...,\gamma_n(t)), \dot\gamma_1(t),...\dot\gamma_n(t) > = \sum\limits_{i=1}^n \partial_if(\gamma(t))\dot\gamma_i(t) $

For the second one we get:

$d^2/dt^2 f(\gamma(t)) = \sum\limits_{i=1}^n<(\partial_{i1}f(\gamma(t)),...,\partial_{in}f(\gamma(t)),\dot\gamma_1(t),...\dot\gamma_n(t)>\dot\gamma_i(t) + \partial_if(\gamma(t))\dot{\dot\gamma_i}(t) = \sum\limits_{i=1}^n \sum\limits_{j=1}^n \partial_{ij}f(\gamma(t))\dot\gamma_i(t)\dot\gamma_j(t) + \sum\limits_{i=1}^n \partial_if(\gamma(t))\dot{\dot\gamma_i}(t) = \dot{\gamma}(t)^t H f(\gamma(t))\dot{\gamma}(t)+ <d/dt f(\gamma(t)),\dot{\dot{\gamma}}(t)> $

Where $< . , . >$ denotes the standard scalar product.


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