Right now I'm learning about differentiability of multivariate function and stumbled upon a particular case:
$g = \gamma: I \subset \mathbb{R} \rightarrow \mathbb{R}^m$
$f: \mathbb{R}^m \rightarrow \mathbb{R}$
for which we can say:
$d(f \circ \gamma)(t_0) = df(\gamma(t_0))d\gamma(t_0)=\langle\nabla f(\gamma(t_0)), \gamma^{\prime}(t_0)\rangle$
Now, if I have for example:
$\gamma(t)= (\begin{array}{} 3t^2+4t\\ \ \ \ \ \ 2t\\ \end{array})$
$f(t_1, t_2) = 2t_1^2+4t_2$
how would I take the composite? And am I even picking the right examples? Thanks for any help.
If I understand correctly, by concatenation you mean the function $$g:I\subset\mathbb{R}\rightarrow\mathbb{R}$$ $$g(t) =f(\gamma(t))$$ In the case my assumption is true, for your particular example, we shall calculate as follows: $$f(\gamma(t))=f{t_{1} \choose t_{2}}=f{3t^2 \choose 2t}=2\times(3t^{2}+4t)^2+4\times(2t)$$
Now, I would imagine you would like to verify that indeed: $$\frac{d}{dt}f(\gamma(t))\bigg\vert_{t=t_{0}}=\langle\nabla f(t)\bigg\vert_{\gamma(t_0)}, \frac{d}{dt}\gamma(t)\bigg\vert_{t=t_{0}}\rangle$$ The above inequality indeed holds, as you can calculate on your own from here. given the following standard calculations: $$\nabla f(t)\bigg\vert_{\gamma(t_0)}= {4t_{1}\choose 4}\bigg\vert_{\gamma(t_0)}={4\times(3t_0^2+4t_0)\choose4}$$ $$\frac{d}{dt}\gamma(t)\bigg\vert_{t=t_{0}}={\frac{d}{dt}(3t^2+4t)\choose \frac{d}{dt}(2t)}\bigg\vert_{t=t_{0}}={6t_0+4\choose2}$$
From here, it is only verification.