I have been trying to solve the following problem:
Verify the chain rule for $f:\mathbb R\rightarrow\mathbb R^2$ defined by $$f(t) =(t,t^2)$$ and
$g: \mathbb R^2\rightarrow\mathbb R^3$ defined by $$g(x_1, x_2)= (x_1^2, x_1x_2, x_2^2-x_1^2)^T$$
I solved this a set under:
$$f '(t) =(1,2t)^T$$
$$g(f(t))= g(t, t^2) = (t^2,t^3, t^4-t^2)^T$$ and hence $$g'(f(t))=(2t, 3t^2, 4t^3-2t)^T$$
I also calculated
$$g'(x_1, x_2) = \left(\begin{matrix}\frac{\partial x_1^2}{\partial x_1} &&\frac{\partial x_1^2}{\partial x_2}\\\frac{\partial x_1x_2}{\partial x_1}&& \frac{\partial x_1x_2}{\partial x_2}\\\frac{\partial (x_2^2-x_1^2)}{\partial x_1}&& \frac{\partial (x_2^2-x_1^2)}{\partial x_2}\end{matrix}\right)$$ which is equal to $$\left(\begin{matrix}2x_1&&0\\x_2&&x_1\\-2x_1&&2x_2\end{matrix}\right)$$
Now I can prove $g'(x_1x_2) f '(t)= g'(f(t))$:
$$\left(\begin{matrix}2x_1&&0\\x_2&&x_1\\-2x_1&&2x_2\end{matrix}\right)\left(\begin{matrix}1\\2t\end{matrix}\right)=\left(\begin{matrix}2x_1\\x_2+2x_1t\\-2x_1+4x_2t\end{matrix}\right)$$
Substituting x$_1$=t and x$_2$=t$^2$ we get
$$\left(\begin{matrix}2t\\3t^2\\-2t+4t^3\end{matrix}\right)$$
which is same as $(2t, 3t^2, 4t^3-2t)^T$
Hence proved.
I am not sure if the procedure is correct .... can we replace $x_1$ and $x_2$ with $t$ and $t^2$ as I have done? Request help with the correct rigorous steps