Let $C^\infty \ni f:\mathbb{R}^3 \to \mathbb{R}$ And $\xi:\mathbb{R}\to\mathbb{R}$ defined by $\xi(t) = f(t,t^2,t^3)$. I want to find $\xi''(t)$.
I defined $g :\mathbb{R} \to \mathbb{R}^3$ as $g(x)=(x,x^2,x^3)$. Calculation of first order derivative is pretty straigh forward using the chain rule, we have :
$D_\xi(t)=D_{f\circ g}(t)=D_f(g(t))D_g(t)=\partial_xf(g(t))+2\partial_yf(g(t))t+3\partial_zf(g(t))t^2$.
However, I have no idea how to differentiate this equation again using the chain rule.
My attempt was :
$\frac{d}{dt}(\partial_xf(g(t))=D_{\partial_xf}(g(t))D_g(t)=\partial_{xx}f(g(t))+2\partial_{xy}f(g(t))t+3\partial_{xz}f(g(t))t^2$
I`m not sure if I applied the chainrule correctly for $\partial_xf$.
Moreover, I tried working with the Hessian matrix, I assumed the value of $\xi''(t)$ would be $D_g(t)H_f(g(t))D_{D_g(x)}(t)$ Where $D_{D_g(x)}(t)= [0,2,6t]$, and $D_g(t)=[1,2t,3t^2]^T$.
I couldn't prove it however,I'll be glad if someone knows of a way of working with the Hessian in this situation.(I bet it`d be much simpler.)
We have $\xi(t)=f\bigl({\bf r}(t))$, where ${\bf r}(t):=(t,t^2,t^3)$. Then $$\eqalign{\xi'(t)&=f_x\bigl({\bf r}(t)\bigr)x'(t)+f_y\bigl({\bf r}(t)\bigr)y'(t)+f_z\bigl({\bf r}(t))z'(t)\cr &=f_x\bigl({\bf r}(t)\bigr)\cdot 1+f_y\bigl({\bf r}(t)\bigr)\cdot 2t+f_z\bigl({\bf r}(t))\cdot 3t^2\ .\cr}$$ For the second derivative we have to use the chain rule again, and also the product rule. I shall omit writing $\bigl({\bf r}(t)\bigr)$ all the time. We obtain $$\eqalign{\xi''(t)&=\bigl(f_{xx}\cdot 1+f_{xy}\cdot 2t+f_{xz}\cdot 3t^2\bigr)\cdot1\cr &\ +\bigl(f_{yx}\cdot1+f_{yy}\cdot2t+f_{yz}\cdot3t^2\bigr)\cdot2t+f_y\cdot2\cr &\ +\bigl(f_{zx}\cdot1+f_{zy}\cdot2t+f_{zz}\cdot3t^2\bigr)\cdot3t^2+f_z\cdot 6t\ ,\cr}$$ which may be somewhat simplified by collecting terms and noting that $f_{xy}=f_{yx}$, etc.