# Why is the derivative the tangent vector?

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?

• @user108205, how would you yourself define or describe the tangent at the point $t$? To help you understand why one thing is the same as another, it helps to know how you understand both things. – Steve Kass Dec 7 '13 at 16:50
• @SteveKass in the geometrical way, taking the example above, $\alpha'(t)$ is a line passing through only one point in the curve $\alpha(t)$ – user108205 Dec 7 '13 at 16:54
• There are many lines that pass "through only one point in the curve $\alpha(t)$." For example, if $\alpha(t)=(a\cos t,a\sin t,bt)$, nearly every straight line passing through the point $(1,0,0)$ passes through only one point on the curve. (And for what it's worth, the tangent line to a curve in general does not necessarily pass through just one point on the curve.) What definition of understanding do you have that deems a particular line among the many through $\alpha(t_0)$ to be the "tangent line" at $t_0$? – Steve Kass Dec 8 '13 at 3:44
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).