<s>[INST] Why is the derivative of Vector equal to Derivative of its rectilinear components? Take a vector $\mathbf A=t^4\mathbf i +t^2\mathbf j$, and call the unit vector along direction of $\mathbf A$ is $\mathbf k$, so the magnitude of this vector $\mathbf A$ along $\mathbf k$ will be $\sqrt{t^8+t^4}$ and thus the vector will be $\sqrt{t^8+t^4}\ \mathbf k$.
So why is its derivative $d\mathbf A/dt=4t^3\mathbf i+2t\mathbf j$? will its magnitude be the same as that of the magnitude of derivative of the vector along $\mathbf k$? How can we distribute its derivative along $\mathbf i$ and $\mathbf j$ and how to prove that their magnitude is same as the magnitude of the derivative of the vector along $\mathbf k$?
 [/INST] After @gandalf61's answer above, the following is just tautology, but let's do it just for the sake of the question.
$$\begin{align}
\mathbf A &=t^4\mathbf i +t^2\mathbf j \\
\end{align}$$
Or, in the direction of $\mathbf A$,
$$\begin{align}
\mathbf A &= |A|.\mathbf k \\
\end{align}$$
Where $\mathbf k$ is the unit vector in the direction of $\mathbf A.$
Thus:
$$\begin{align}
\mathbf A &= \sqrt{t^8+t^4}.\frac{t^4\mathbf i +t^2\mathbf j}{\sqrt{t^8+t^4}} \\
\end{align}$$
Now, we can derive $\frac{d\mathbf A}{dt}$:
$$\begin{align}
\frac{d\mathbf A}{dt} &= \frac{d|A|}{dt}.\mathbf k + |A|.\frac{d\mathbf k}{dt} \\
&=\frac{8t^7 + 4t^3}{2\sqrt{t^8+t^4}}.\frac{t^4\mathbf i +t^2\mathbf j}{\sqrt{t^8+t^4}} + \sqrt{t^8+t^4}.\frac{2t^7 \mathbf i - 2t^9\mathbf j}{(t^8+t^4)^{3/2}} \\
&= \frac{(4t^{11} + 2t^7 + 2t^7) \mathbf i + (4t^9 + 2t^5 -2t^9)\mathbf j}{t^8+t^4} \\
&= \frac{t^4[(4t^7 + 4t^3)\mathbf i + (2t^5 + 2t) \mathbf j]}{t^4(t^4+1)} \\
&= \frac{(t^4+1)(4t^3 \mathbf i + 2t\mathbf j)}{t^4+1} \\
&= 4t^3 \mathbf i + 2t\mathbf j
\end{align}$$
So, we see that the derivative of $\mathbf A$ is indeed the same whether we represent $\mathbf A$ in terms of $\mathbf i$ and $\mathbf j$, or in terms of $\mathbf k$ (and subsequently $\mathbf k$ in terms of $\mathbf i$ and $\mathbf j$).
</s>