求dy/dx并化简各项系数：求导结果验证及疑问咨询

Hey there! Let's work through your questions one by one:

First off, great job—your derivative calculation is completely correct! Let's verify it step by step to be 100% sure:
Given the original function:
$y = 5x - 7 + 2x^{-1/2} + 3x^{-1}$
We apply the power rule ($\frac{d}{dx}(x^n) = nx^{n-1}$) to each term:

• The derivative of $5x$ is $5$ (since $1 \cdot 5x^{0} = 5$)
• The derivative of the constant -7 is $0$
• For $2x^{-1/2}$: $2 \cdot (-\frac{1}{2})x^{-\frac{1}{2}-1} = -x^{-\frac{3}{2}}$
• For $3x^{-1}$: $3 \cdot (-1)x^{-1-1} = -3x^{-2}$
  Adding these up gives exactly your result: $\frac{dy}{dx} = 5 - x^{-\frac{3}{2}} - 3x^{-2}$

You're right to think your coefficients are already in simplest form! This instruction usually applies to cases where coefficients can be simplified further—like if you ended up with a fraction that can be reduced (e.g., $\frac{6}{2}x^3$ becomes $3x^3$), a mixed number converted to an improper fraction, or a radical that needs rationalizing. Since your coefficients are all integers (-1, -3, 5) with no room for further simplification, your result fully meets this requirement.

Even though the steps feel short, writing out each individual step explicitly will help you avoid losing points. Here's how to structure it for clarity:

• Start by restating the original function clearly
• Break down the derivative of each term, showing the application of the power rule (even if it seems obvious)
• Combine the results to get the final derivative

For example, your work could look like this:

Given $y = 5x - 7 + 2x^{-1/2} + 3x^{-1}$, we find $\frac{dy}{dx}$ by differentiating term-by-term:

• $\frac{d}{dx}(5x) = 5$
• $\frac{d}{dx}(-7) = 0$
• $\frac{d}{dx}(2x^{-1/2}) = 2 \cdot (-\frac{1}{2})x^{-\frac{3}{2}} = -x^{-\frac{3}{2}}$
• $\frac{d}{dx}(3x^{-1}) = 3 \cdot (-1)x^{-2} = -3x^{-2}$

Adding these together: $\frac{dy}{dx} = 5 - x^{-\frac{3}{2}} - 3x^{-2}$

This shows graders that you understand the underlying rules, not just that you got the right answer.

内容的提问来源于stack exchange，提问作者Ben Hughes