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

(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,


*

*Apply the Product Rule on $h$ to get $h' = f'g + fg'$.

*Apply the Chain Rule on both $f$ and $g$.


However, I can't find a trick neater than Feynman's one.
A: hint
$$h^3 (t)=g (t)=(t+1)^2 (2t^2-1)^9 $$
$$3h^2 (t)h'(t)=g'(t) $$
$$=2(2t^2-1)^8 (t+1)(20t^2+18t-1)$$
$$h'(t)=\frac 23 (2t^2-1)^2 (t+1)^{-\frac 13}(20t^2+18t-1) $$
