Differentiate using chain and product rule, then simplify. $$h(t)= (t+1)^{2/3}(2t^2-1)^3$$
Differentiate the above expression.
First, I applied product and chain rules:
$$h'(t)=(t+1)^{2/3}3(2t^2-1)^2(4t)+\frac{2}{3}(t+1)^{-1/3}(2t^2-1)^3$$
At first I was satisfied with this answer, as it's quite difficult to simplify this expression. However, the textbook provides the following, more simplified answer:
$$h'(t)=\frac{2}{3}(t+1)^{-1/3}(2t^2-1)^2(20t^2+18t-1)$$
Assuming my expression is correct, how did they reach this solution? (For what it's worth, I do see a couple of my terms agree with the answer). 
For a student in higher level calculus courses (i.e. Multivariable Calculus, Calc III), should it be a given that this simplification should be obvious? Thanks in advance for your help.
 A: Your first steps were correct.
$$(t+1)^{2/3}3(2t^2-1)^2(4t)+\frac{2}{3}(t+1)^{-1/3}(2t^2-1)^3$$
$$=\left(\frac{2}{3}(t+1)^{-1/3}(2t^2-1)^2\right)\frac{3}{2}(3\cdot 4\cdot t)(t+1)+$$
$$+\left(\frac{2}{3}(t+1)^{-1/3}(2t^2-1)^2\right)\cdot (2t^2-1)$$
$$=\left(\frac{2}{3}(t+1)^{-1/3}(2t^2-1)^2\right)$$
$$\cdot \left(18t(t+1)+2t^2-1\right)$$
$$=\frac{2}{3}(t+1)^{-1/3}(2t^2-1)^2(20t^2+18t-1)$$
I used the distributive property on $\left(\frac{2}{3}(t+1)^{-1/3}(2t^2-1)^2\right)$.
I hope this is clear.
A: \begin{align}
&(t+1)^{2/3}3(2t^2-1)^2(4t)+\frac{2}{3}(t+1)^{-1/3}(2t^2-1)^3\\
=&\color{red}{\frac{2}{3}(t+1)^{-1/3}(2t^2-1)^2}(t+1)(18t)+\color{red}{\frac{2}{3}(t+1)^{-1/3}(2t^2-1)^2}(2t^2-1)\\
=&\color{red}{\frac{2}{3}(t+1)^{-1/3}(2t^2-1)^2}\left((t+1)(18t)+2t^2-1\right)\\
=&\frac{2}{3}(t+1)^{-1/3}(2t^2-1)^2(20t^2+18t-1)
\end{align}
It's a simple matter of factorization.
A: Hint
When you face combinations of products, quotients, powers, logarithmic differentiation makes life easier $$h= (t+1)^{\frac 23}(2t^2-1)^3\implies \log(h)=\frac 23\log(t+1)+3\log(2t^2-1)$$ Differentiate both sides $$\frac{h'}h=\frac 23 \frac 1{t+1}+3\frac {4t}{2t^2-1}=\frac{40 t^2+36 t-2}{3 (t+1) \left(2 t^2-1\right)}$$ Now $$h'=h \times \frac{h'}h=(t+1)^{\frac 23}(2t^2-1)^3\times \frac{40 t^2+36 t-2}{3 (t+1) \left(2 t^2-1\right)}$$ Simplify and you are done.
