How to derive this equation. Combo of chain rule and product rule

Question

I'm curious, why can't you use chain rule first? Why does product rule have to come first?

$$h(t) = (t+1)^{2/3} \cdot (2t^2-1)^3$$ $$h'(t) = (t+1)^{2/3} \cdot (3(2t^2-1)^2 \cdot 4t) + (2t^2-1)^3\cdot \frac{2}{3}(t+1)^{-1/3} \cdot 1$$ $$h'(t) = (t+1)^{2/3} \cdot (12t(2t^2-1)^2) + (2t^2-1)^3 \cdot \frac{2}{3}(t+1)^{-1/3}$$

How do I simplify from here? I'm not too sure how to pull out that $(t+1)$ since there is a mix of positive and negative integers. Where do I go from here?

Answer 1

With no existing answers with positive score, I expand my comment into an answer.

$$ \begin{aligned} h(t) &= (t+1)^{2/3} (2t^2-1)^3 \\ h'(t) &= h(t) \left[ \frac23 \frac{1}{t+1} + 3 \frac{4t}{2t^2 - 1} \right] \\ &= (t+1)^{2/3} (2t^2-1)^3 \cdot \left[ \frac23 \cdot \frac{1}{t+1} + 3 \cdot \frac{4t}{2t^2 - 1} \right] \\ &= \frac23 (t+1)^{-1/3} (2t^2-1)^3 + 12t (t+1)^{2/3} (2t^2-1)^2 \quad\text{(This line is redundant.)} \\ &= \frac23 (t+1)^{-1/3} (2t^2-1)^2 \left[ (2t^2-1) + 18t(t+1) \right] \\ &= \frac23 (t+1)^{-1/3} (2t^2-1)^2 (20t^2 + 18t -1) \end{aligned} $$

Remarks: I have left a redundant line in the answer to show that my calculations are consistent with the question body.

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(Edit in response to OP's doubts.)

Why can't you use chain rule first? Why does product rule have to come first?

I would rather ask how to apply chain rule first in $$(f \circ g)'(x) = f'(g(x)) g'(x).$$ What $f$ and $g$ can be chosen?

The Chain Rule is applied on composite functions, but first $h$ is a product of two functions $f(t) = (t+1)^{2/3}$ and $g(t) = (2t^2 - 1)^3$. Observe that both $f$ and $g$ are composite functions. Consequently,

1. Apply the Product Rule on $h$ to get $h' = f'g + fg'$.
2. Apply the Chain Rule on both $f$ and $g$.

However, I can't find a trick neater than Feynman's one.

Answer 2

hint

$$h^3 (t)=g (t)=(t+1)^2 (2t^2-1)^9 $$

$$3h^2 (t)h'(t)=g'(t) $$

$$=2(2t^2-1)^8 (t+1)(20t^2+18t-1)$$

$$h'(t)=\frac 23 (2t^2-1)^2 (t+1)^{-\frac 13}(20t^2+18t-1) $$