幂法则进阶证明请求：有理数指数函数导数推导

Let's build on the results you already have to prove the power rule for rational exponents, step by step.

First, a quick recap of what's already confirmed:

• Using induction and the product rule, we've shown for all positive integers $n \in \mathbb{N}$, $\frac{d}{dx}(x^n) = nx^{n-1}$
• Via the quotient rule, we proved for all positive integers $n \in \mathbb{N}$, $\frac{d}{dx}(x^{-n}) = -nx^{-n-1}$

Now let's tackle the remaining cases.

1. Prove $\frac{d}{dx}(x^{1/n}) = \frac{1}{n}x^{\frac{1}{n}-1}$ for all $n \in \mathbb{Z} \setminus {0}$ #

Let's start by letting $y = x^{1/n}$. By definition, this means $y^n = x$ (raising both sides to the nth power cancels out the root). Now we'll use implicit differentiation (which relies on the chain rule, as suggested).

Differentiate both sides with respect to $x$:

• Left side: Applying the chain rule to $y^n$ (since $y$ is a function of $x$), we get $\frac{d}{dx}(y^n) = n y^{n-1} \cdot \frac{dy}{dx}$
• Right side: The derivative of $x$ with respect to $x$ is simply $1$

Now solve for $\frac{dy}{dx}$:
$$
n y^{n-1} \cdot \frac{dy}{dx} = 1 \
\frac{dy}{dx} = \frac{1}{n y^{n-1}}
$$

Substitute back $y = x^{1/n}$ to get the derivative in terms of $x$:
$$
\frac{dy}{dx} = \frac{1}{n (x^{1/n}){n-1}} = \frac{1}{n x^{\frac{n-1}{n}}} = \frac{1}{n x^{1 - \frac{1}{n}}} = \frac{1}{n} x^{\frac{1}{n} - 1}
$$

This works for all non-zero integers $n$ (positive or negative) as long as $x$ is in the domain where $x^{1/n}$ is defined.

2. Prove $\frac{d}{dx}(x^{p/q}) = \frac{p}{q}x^{\frac{p}{q}-1}$ for all $\frac{p}{q} \in \mathbb{Q}$ #

Any rational number can be written as $\frac{p}{q}$ where $p \in \mathbb{Z}$ and $q \in \mathbb{N} \setminus {0}$. Let's rewrite $x^{p/q}$ as $(x^{1/q})p$—this lets us use the chain rule with the results we already have.

Let $u = x^{1/q}$. Our function becomes $u^p$. Applying the chain rule:
$$
\frac{d}{dx}(u^p) = p u^{p-1} \cdot \frac{du}{dx}
$$

From step 1, we know $\frac{du}{dx} = \frac{1}{q}x^{\frac{1}{q}-1}$. Substitute back $u$ and $\frac{du}{dx}$:
$$
\frac{d}{dx}(x^{p/q}) = p (x^{1/q}){p-1} \cdot \frac{1}{q}x^{\frac{1}{q}-1}
$$

Now simplify the exponents:

• $(x^{1/q}){p-1} = x^{\frac{p-1}{q}}$
• Adding the exponents when multiplying: $\frac{p-1}{q} + \frac{1}{q} - 1 = \frac{p}{q} - 1$

Putting it all together gives us:
$$
\frac{d}{dx}(x^{p/q}) = \frac{p}{q} x^{\frac{p}{q} - 1}
$$

That's the power rule for all rational exponents!

内容的提问来源于stack exchange，提问作者Brandon O'Neil