# 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.

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.

\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.

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.