Prove the power rule for positive integral exponents n by writing y=u^{n} , expanding \Delta y=\left( u+\Delta u\right) ^{n}-u^{n} by the binomial theorem, and then dividing by \Delta x . Use the quotient rule to extend this result to negative integral exponents.

给出两个来自chatgpt的参考证明方法，显然方法二更符合题意（可以直接看方法二）。

方法一：

Let y=u^n , where n is a positive integer.

Expanding \Delta y = (u + \Delta u)^n - u^n using the binomial theorem, we get:

\Delta y = u^n + \binom{n}{1}u^{n-1}(\Delta u) + \binom{n}{2}u^{n-2}(\Delta u)^2 + \cdots + \binom{n}{n-1}u(\Delta u)^{n-1} + (\Delta u)^n - u^n

Simplifying, we can see that all the terms containing powers of (\Delta u) greater than or equal to 2 will have a factor of \Delta u , so dividing by \Delta x and taking the limit as \Delta u approaches 0, those terms will disappear. Thus, we get:

\lim_{\Delta u \to 0} \frac{\Delta y}{\Delta u} = \lim_{\Delta u \to 0} \left(nu^{n-1} + \binom{n}{1}u^{n-2}(\Delta u) + \cdots + \binom{n}{n-2}(\Delta u)^{n-2}u + (\Delta u)^{n-1}\right)

= nu^{n-1}

Therefore, we have shown that \frac{d}{dx}(u^n) = nu^{n-1} for positive integral exponents n.

To extend this result to negative integral exponents, we can use the quotient rule. Let y = \frac{1}{u^n} . Then, using the power rule that we just proved, we have:

\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{u^n}\right) = -\frac{n}{u^{n+1}}

Multiplying both sides by -1 , we get:

\frac{d}{dx}(u^{-n}) = (-n)u^{-n-1}

Thus, we have shown that the power rule holds for all integer exponents.

方法二：

To prove the power rule for positive integral exponents n, we start by writing y=u^n , where u is a function of x. Let us now apply the binomial theorem to expand \Delta y=(u+\Delta u)^n - u^n ：

\Delta y=(u+\Delta u)^n - u^n = \sum_{k=0}^{n} {n\choose k}u^{n-k}(\Delta u)^k

where {n\choose k}=\frac{n!}{k!(n-k)!} is the binomial coefficient.

Now, dividing by \Delta x yields:

\frac{\Delta y}{\Delta x} = \frac{\sum_{k=0}^{n} {n\choose k}u^{n-k}(\Delta u)^k}{\Delta x}

Using the limit definition of the derivative, we can let \Delta x approach zero:

\frac{dy}{dx} = \lim_{\Delta x\to 0} \frac{\sum_{k=0}^{n} {n\choose k}u^{n-k}(\Delta u)^k}{\Delta x}

We can simplify the numerator by factoring out (\Delta u) ：

\frac{dy}{dx} = \lim_{\Delta x\to 0} \frac{\Delta u}{\Delta x}\sum_{k=0}^{n} {n\choose k}u^{n-k}(\Delta u)^{k-1}

As \Delta x approaches zero, \Delta u/\Delta x approaches du/dx , and (\Delta u)^{k-1} approaches zero for k\geq 2 （连续函数的性质）. Therefore, we can simplify the expression to:

\frac{dy}{dx} = nu^{n-1}\frac{du}{dx}

This is the power rule for positive integral exponents.

To extend this result to negative integral exponents, we can use the quotient rule. Let n be a negative integer and y=u^{-n} , then:

\frac{dy}{dx} = \frac{d}{dx}\left(u^{-n}\right) = \frac{d}{dx}\left(\frac{1}{u^n}\right)

Using the quotient rule:

\frac{dy}{dx} = -\frac{nu^{n-1}}{u^{2n}}\frac{du}{dx} = -n\frac{du}{dx}u^{-n-1}

Thus, the power rule holds for all integer exponents.

Prove the power rule for positive integral exponents n by writing y=u^{n} , expanding \Delta y=\left( u+\Delta u\right) ^{n}-u^{n} by the binomial theorem, and then dividing by \Delta x . Use the quo­tient rule to extend this result to negative integral expo­nents.

Prove the power rule for positive integral exponents n by writing y=u^{n} , expanding \Delta y=\left( u+\Delta u\right) ^{n}-u^{n} by the binomial theorem, and then dividing by \Delta x . Use the quotient rule to extend this result to negative integral exponents.