基于第一性原理及泰勒级数展开证明e^x的导数为e^x

Let's work through these two calculus problems step by step—they're classic results, but breaking them down helps solidify the fundamentals.

1. Derivative of (e^x) via First Principles #

First, recall the core definition of a derivative from first principles:
[
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
]
For our function (f(x) = e^x), substitute into the formula:
[
\frac{d}{dx}e^x = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h}
]
Since (e^{x+h} = e^x \cdot e^h), we can factor out (e^x) (it doesn't depend on (h), so it stays outside the limit):
[
= e^x \cdot \lim_{h \to 0} \frac{e^h - 1}{h}
]
Now we just need to evaluate (\lim_{h \to 0} \frac{e^h - 1}{h}). Remember that (e) is defined as (\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n), which means (e^h = \lim_{n \to \infty} \left(1 + \frac{h}{n}\right)^n). Using the binomial theorem to expand this:
[
\left(1 + \frac{h}{n}\right)^n = 1 + n \cdot \frac{h}{n} + \frac{n(n-1)}{2!} \cdot \left(\frac{h}{n}\right)^2 + \frac{n(n-1)(n-2)}{3!} \cdot \left(\frac{h}{n}\right)^3 + \dots
]
Simplify each term as (n \to \infty) (terms like (\frac{n-1}{n}) approach 1):
[
e^h = 1 + h + \frac{h^2}{2!} + \frac{h^3}{3!} + \dots
]
Plug this back into our limit:
[
\lim_{h \to 0} \frac{e^h - 1}{h} = \lim_{h \to 0} \frac{\left(1 + h + \frac{h^2}{2!} + \dots\right) - 1}{h} = \lim_{h \to 0} \left(1 + \frac{h}{2!} + \frac{h^2}{3!} + \dots\right)
]
As (h) approaches 0, all terms with (h) vanish, leaving just 1. So:
[
\frac{d}{dx}e^x = e^x \cdot 1 = e^x
]

2. Proving (d/dx e^x = e^x) Using Taylor Series + First Principles #

First, write out the Taylor series (Maclaurin series, centered at 0) for (e^x), keeping at least the first 4 terms to make the algebra clear:
[
e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \sum_{k=4}^{\infty} \frac{x^k}{k!}
]
Now expand (e^{x+h}) using the same series, replacing (x) with (x+h):
[
e^{x+h} = 1 + (x+h) + \frac{(x+h)^2}{2!} + \frac{(x+h)^3}{3!} + \sum_{k=4}^{\infty} \frac{(x+h)^k}{k!}
]
Subtract (e^x) from (e^{x+h}) to get the numerator of our first-principles formula:
[
e^{x+h} - e^x = \left[1 + x + h + \frac{x^2 + 2xh + h^2}{2} + \frac{x^3 + 3x^2h + 3xh^2 + h^3}{6} + \dots\right] - \left[1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots\right]
]
Cancel out the matching terms (1, (x), (\frac{x^2}{2}), (\frac{x^3}{6})) and simplify the rest:
[
= h + xh + \frac{h^2}{2} + \frac{x^2h}{2} + \frac{xh^2}{2} + \frac{h^3}{6} + \text{higher-order terms with } h
]
Factor out an (h) from every term:
[
= h \left[1 + x + \frac{x^2}{2} + \frac{h}{2} + \frac{xh}{2} + \frac{h^2}{6} + \text{higher-order terms with } h\right]
]
Divide by (h) (since (h \neq 0) when taking the limit):
[
\frac{e^{x+h} - e^x}{h} = 1 + x + \frac{x^2}{2} + \frac{h}{2} + \frac{xh}{2} + \frac{h^2}{6} + \text{higher-order terms with } h
]
Now take the limit as (h \to 0): all terms containing (h) go to 0, leaving us with:
[
\lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = 1 + x + \frac{x^2}{2} + \dots
]
But that's exactly the Taylor series for (e^x)! So we've shown:
[
\frac{d}{dx}e^x = e^x
]

内容的提问来源于stack exchange，提问作者K.Faz