# Equivalence of the definition of differentials

I have two definitions of a differential of a map $$f: \mathbb{R}^n \rightarrow \mathbb{R}$$ at $$x$$ denoted as $$df_x$$.

1. $$df_x(h) = \sum_{i = 1}^n \frac{\partial f (x)}{\partial x_i}h_i$$
2. $$df_x(h) = \lim_{t \rightarrow 0} \frac{f(x + ht) - f(x)}{t}$$

How do I go about showing this equivalence? Intuitively the equivalence makes sense. If we take $$n = 2$$, the partials in each direction give a sense of how much $$f$$ changes in each respective dimension, so (1) makes sense. I can also see how (2) gives us the differential if we think of the change of a surface in the direction of $$h$$.

I started with (2) and began by saying $$df_x(h) + E(t) = \frac{f(x + ht) - f(x)}{t}$$ such that $$\lim_{t \rightarrow 0} E(t) = 0$$. This gives us $$f(x + h) = f(x) + df_x(h)t + o(t)$$. I then tried to show that $$f(x + h) = f(x) + (\sum_{i = 1}^n \frac{\partial f(x)} + {\partial x_i}h_i) + o(t)$$ but I'm not sure if this helps and I am stuck here.

• There's a crucial piece of information missing: The differential is linear as a function of $h$. Does that help? May 14 at 13:19
• @AndrewD.Hwang - Thank you for the response. How do you know that it is linear? It is clear from (1) but not from (2). Is there a way we can show linearity with (2)? 2 days ago
• In this setting, $f$ mapping is differentiable at $x$ if there exists a linear transformation $df(x)$ such that $$f(x+h)=f(x)+df(x)(h)+o(|h|).$$This cannot be deduced this algebraically from your conditions, but must be assumed. 2 days ago