复合乘积函数求导疑问：为何先用乘积法则？导数如何化简？

Why We Use the Product Rule First (Not the Chain Rule) #

Great question—let’s clear up the order of operations here. The function (h(t)) is a product of two distinct composite functions:

• Let (u(t) = (t+1)^{2/3}) (a composite function: outer power function (x^{2/3}), inner linear function (t+1))
• Let (v(t) = (2t^2-1)3) (another composite function: outer power function (x^3), inner quadratic function (2t^2-1))

The Chain Rule is designed for single nested functions (i.e., (f(g(t))), where one function wraps around another). But (h(t)) isn’t a single nested function—it’s the multiplication of two separate functions. That’s exactly when the Product Rule ((\frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t))) needs to come first. We only bring in the Chain Rule after applying the Product Rule, to calculate (u'(t)) and (v'(t)) (since each of those is a composite function on its own).

If you tried to use the Chain Rule first, you’d be forcing (h(t)) into a shape it doesn’t fit—there’s no way to write (h(t)) as a single (f(g(t))), so the Chain Rule doesn’t apply to the entire product.

Simplifying Your Initial Derivative #

You already nailed the starting point:
$$h'(t) = (t+1)^{2/3} \cdot 12t(2t^2-1)2 + (2t^2-1)3 \cdot \frac{2}{3}(t+1)^{-1/3}$$

Here’s how to factor out common terms and simplify step by step:

1. Spot the lowest common factors across both terms:

   • For ((t+1)): the smallest exponent is (-1/3) (since (2/3 > -1/3))
   • For ((2t^2-1)): the smallest exponent is (2) (since (3 > 2))
     The shared factor we can pull out is ((2t^2-1)2 \cdot (t+1)^{-1/3})

2. Rewrite each term to expose this common factor:

   • First term: ((t+1)^{2/3} \cdot 12t(2t^2-1)2 = (t+1)^{-1/3} \cdot (t+1)^1 \cdot 12t(2t^2-1)2) (because (-1/3 + 1 = 2/3))
   • Second term: ((2t^2-1)3 \cdot \frac{2}{3}(t+1)^{-1/3} = (2t^2-1)2 \cdot (2t^2-1)1 \cdot \frac{2}{3}(t+1)^{-1/3}) (because (2 + 1 = 3))

3. Factor out the common term:
   $$
   h'(t) = (2t^2-1)2 \cdot (t+1)^{-1/3} \left[ 12t(t+1) + \frac{2}{3}(2t^2-1) \right]
   $$

4. Simplify the expression inside the brackets:

   • Get a common denominator (3) to combine the terms:
     $$
     12t(t+1) = \frac{36t(t+1)}{3}
     $$
   • Expand and combine the numerators:
     $$
     36t(t+1) + 2(2t^2-1) = 36t^2 + 36t + 4t^2 - 2 = 40t^2 + 36t - 2
     $$
   • Factor out a 2 from the numerator to clean it up:
     $$
     40t^2 + 36t - 2 = 2(20t^2 + 18t - 1)
     $$

5. Combine everything into a polished form:
   Substitute back in, and optionally rewrite ((t+1)^{-1/3}) as (\frac{1}{(t+1)^{1/3}}) for a positive exponent:
   $$
   h'(t) = \frac{(2t^2-1)2 \cdot 2(20t^2 + 18t - 1)}{3(t+1)^{1/3}}
   $$
   Or if you prefer keeping negative exponents:
   $$
   h'(t) = \frac{2}{3}(2t^2-1)2(t+1)^{-1/3}(20t2 + 18t - 1)
   $$

内容的提问来源于stack exchange，提问作者Jwan622