Differential as a tangent to $\Delta f(x)$

Question

I had come across the following

The number $h = x − x_0$ , that is, the increment of the argument, can be regarded as a vector attached to the point $x_0$ and defining the transition from $x_0$ to $x = x_0 + h$. We denote the set of all such vectors by $ T\mathbb R(x_0)$. It can then be seen that the mapping $$df(x_0):T\mathbb R(x_0) \rightarrow T\mathbb R(f(x_0))$$ defined by the differential $h\mapsto f'(x_0)h$ is tangent to the mapping $$h \mapsto f(x_0 + h) - f(x_0)$$

This might be a trivial question but I was wondering how would we go about proving this? Would not the fact that this is a tangent to the change in f indicate that we need to differentiate it wrt h but how would that make $df(x_0)$ a tangent?

Answer 1

$\newcommand{\Reals}{\mathbf{R}}$Suppose $f$ is a real-valued function defined in a neighborhood of a real number $x_{0}$. Two standard definitions of $f$ being _differentiable_ at $x_{0}$ are

1. The limit of the difference quotients exists: $$ f'(x_{0}) := \lim_{h \to 0} \frac{f(x_{0} + h) - f(x_{0})}{h}\quad\text{exists.} $$
2. There exists a real number $f'(x_{0})$ such that the function $E$ defined in a neighborhood of $0$ by $$ f(x_{0} + h) = f(x_{0}) + f'(x_{0})h + E(h) $$ satisfies $\lim\limits_{h \to 0} \dfrac{E(h)}{h} = 0$.

These are equivalent since $$ f'(x_{0}) = \lim_{h \to 0} \frac{f(x_{0} + h) - f(x_{0})}{h} $$ if and only if \begin{align*} 0 &= \lim_{h \to 0} \frac{f(x_{0} + h) - f(x_{0})}{h} - f'(x_{0}) \\ &= \lim_{h \to 0} \frac{f(x_{0} + h) - f(x_{0}) - hf'(x_{0})}{h}. \end{align*} The point for this question is, it's the same function $f$ throughout.

In other words, define functions $g$ and $L = df(x_{0})$ in some neighborhood of $0$ by $$ g(h) = f(x_{0} + h) - f(x_{0}),\qquad L(h) = hf'(x_{0}). $$ Geometrically, if we shift the origin to lie at $(x_{0}, f(x_{0}))$, i.e. we set $x = x_{0} + h$, $y = f(x_{0}) + k$, and use $(h, k)$ as Cartesian coordinates, then the Cartesian graph $y = f(x)$ has equation $k = g(h)$ and the tangent line has equation $k = L(h)$.

Analytically, tangency comes down to the difference $|g(h) - L(h)|$ being negligible compared to $|h|$ in the limit as $h \to 0$.

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If the goal is to understand all this on smooth manifolds in a coordinate-invariant way, a bit more care is required. First fix a local coordinate $x$ near $x_{0}$, and let $h$ be the induced coordinate on the tangent space $T_{x_{0}}\Reals$, i.e., identify tangent vectors at $x_{0}$ with $h\, \partial_{x}(x_{0})$. Similarly let $y$ be a coordinate near $f(x_{0})$ and let $k$ be the induced coordinate on $T_{f(x_{0})}\Reals$. In these local trivializations of the tangent bundle, use the preceding discussion to show the graph of $f$ and the graph of its tangent line at $x_{0}$ are tangent at $(x_{0}, f(x_{0}))$. Finally, show that the resulting tangency does not depend on the choice of coordinates.