链式法则与乘积法则结合求导的化简正确性验证及优化建议咨询

Hey there! Let's break down your derivative simplification step by step to catch where things went sideways, and I'll share some practical tips to keep this process smoother in the future.

Initial Derivative Check #

First off, your application of the product rule is perfect! For $f(t) = (3t-1)^4 \cdot (2t + 1)^{-3}$, you correctly applied $\frac{d}{dt}[u \cdot v] = u'v + uv'$:
$$f'(t) = (3t-1)^4 \cdot -3(2t+1)^{-4} \cdot 2 + (2t+1)^{-3} \cdot 4(3t-1)^3 \cdot 3$$
And simplifying the coefficients gives:
$$= -6(3t-1)^4(2t+1){-4} + 12(3t-1)^3(2t+1){-3}$$
No issues here—great job getting to this point!

Where You Went Off Track #

Your factoring step had two small but critical errors:

1. Sign confusion in factoring: When you extracted $-6(3t-1)^3(2t+1){-3}$, the bracket logic was almost right, but the follow-up simplification got tripped up.
2. Typo in the numerator: You wrote $\frac{3t+1}{2t+1}$ instead of $\frac{3t-1}{2t+1}$—that's an easy slip-up when rushing through steps!

Correct Simplification Walkthrough #

Let's redo the factoring and simplification properly:
Start with the correct intermediate derivative:
$$f'(t) = -6(3t-1)^4(2t+1){-4} + 12(3t-1)^3(2t+1){-3}$$

Step 1: Pick the right common factor #

Instead of extracting a negative factor (which can muddle signs), extract the positive greatest common factor of the two terms:

• For $(3t-1)$: take the lowest power, $(3t-1)^3$
• For $(2t+1)$: take the lowest power, $(2t+1)^{-4}$ (since $-4 < -3$)
• Coefficients: GCD of 6 and 12 is 6

This gives us:
$$f'(t) = 6(3t-1)^3(2t+1){-4} \left[ -(3t-1) + 2(2t+1) \right]$$

Step 2: Simplify the bracket #

Calculate the expression inside the brackets:
$$-(3t-1) + 2(2t+1) = -3t + 1 + 4t + 2 = t + 3$$

Step 3: Clean up the final form #

Convert negative exponents to fractions for a cleaner result:
$$f'(t) = \frac{6(3t-1)^{3(t+3)}{(2t+1)}4}$$

Pro Tips for Avoiding Simplification Chaos #

• Stick to positive common factors first: This reduces the chance of sign errors that derail your work.
• Write every step explicitly: Don't skip expanding terms or rewriting exponents—typos like your $3t+1$ mistake happen when you rush.
• Cross-verify with another method: Use logarithmic differentiation to double-check your result:
  1. Take the natural log of $f(t)$: $\ln f(t) = 4\ln(3t-1) - 3\ln(2t+1)$
  2. Differentiate both sides: $\frac{f'(t)}{f(t)} = \frac{12}{3t-1} - \frac{6}{2t+1}$
  3. Multiply by $f(t)$ and simplify—you'll get the same final result, confirming you're on the right track.

内容的提问来源于stack exchange，提问作者Jwan622