I'm trying to understand, at least intuitively why the derivative of a function at a point is the tangent vector at this point.
If we see the functions of this form $f:\mathbb R\to \mathbb R$ we see clearly that
$$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$$
is the slope of the tangent of $f$ at the point $a\in \mathbb R$, because the angular coefficient of a line is $\frac{\Delta y}{\Delta x}$.
However I couldn't understand why the derivative is the tangent vector in higher dimensions so clearly as we see in my example above.
In order to illustrate what I said above, let's take for example the helix
$$\alpha(t):\mathbb R\to \mathbb R^3,\ \alpha(t)=(a\cos t,a\sin t,bt)$$
If we take the definition of derivative in higher dimensions in Spivak's book we have:
A function $f:\mathbb R^n\to \mathbb R^m$ is differentiable at $a\in \mathbb R^m$ if there is a linear transformation $\lambda: \mathbb R^n\to \mathbb R^m$ such that
$$\lim_{h\to 0}\frac{|f(a+h)-f(a)-\lambda(h)|}{|h|}=0$$
I can't see why the $\alpha'(t)$ is the tangent at the point $t$ and I tried also see the derivative as the Jacobian without success.
So my question is why does the derivative is the tangent vector?
Thanks in advance.
Best Answer
In 2 and 3 dimensions we can observe from example that the derivative satisfies what is required of a tangent vector. If you understand why is that it works in the case $f:R \rightarrow R$, then you will get why it works in the case for 2 and 3 dimensions. For higher dimensions this is just a generalization as you cannot visualize 4d space or higher (I can hardly visualize 3d).