Let $\varphi:\mathbb{R}\to\mathbb{R}^n$ and $F:\mathbb{R}^n\to\mathbb{R}$. I'd like to find $D^2(F\circ\varphi)$.
$(D(F\circ\varphi))_t=(DF)_{\varphi(t)}\circ (D\varphi)_t$.
I'm a little confused as to how you would proceed to find the derivative of $D(F\circ\varphi)$. You'd want to think of $ D(F\circ\varphi)$ without the $t$ dependence,so $D(F\circ\varphi)=(DF)_{\varphi}\circ DF$. So then applying the chain rule again, $(D^2(F\circ\varphi))_t=(D(DF\circ\varphi))_{(DF)_t}\circ (D^2F)_t$. This doesn't seem right to me. Can anyone explain?
As the derivative of $G=F∘φ$ is a product $G'=(F'∘φ)\cdot φ'$, you have to use the product rule for the second derivative, $$ G''=(F''∘φ)[φ',φ']+(F'∘φ)\cdot φ''. $$ Note that as linear algebra object, $F''$ is a third order tensor, or a vector valued symmetric bilinear form.