How to find the derivation of this chain rule/product rule comb.Check

Question

I just wanted to make sure that I was simplifying this correctly. Sometimes my simplifcation is messy and I'm not sure why. Any advice?

$$f(t) = (3t-1)^4 \cdot (2t + 1)^{-3} $$

$$f'(t) = (3t-1)^4 \cdot -3(2t+1)^{-4} \cdot 2 + (2t+1)^{-3} \cdot 4(3t-1)^3 \cdot 3$$

$$= (3t-1)^4 \cdot -6(2t+1)^{-4} + (2t+1)^{-3} \cdot 12(3t-1)^3$$

and now pulling out common factors:

$$-6(3t-1)^3(2t+1)^{-3} \cdot [(3t-1)(2t+1)^{-1} + (-2)]$$

$$-6(3t-1)^3(2t+1)^{-3} \cdot [\frac{3t+1}{2t+1} - 2]$$

$$\frac{6(3t-1)^3}{(2t+1)^3} \cdot [2 - \frac{3t+1}{2t+1}]$$

Is that remotely correctly simplified?

Answer 1

Yes it's correct. Although, it can be simplified furthermore -

\begin{align} \frac{6(3t-1)^3}{(2t+1)^3} \cdot \left(\color{blue}{2} - \frac{3t-1}{2t+1}\right)&= \frac{6(3t-1)^3}{(2t+1)^3} \cdot \left(\color{blue}{\frac{2(2t+1)}{2t+1}}-\frac{3t-1}{2t+1}\right) \\ &=\frac{6(3t-1)^3}{(2t+1)^3} \cdot \left(\frac{\color{blue}{(4t+2)}-(3t-1)}{2t+1}\right) \\ &=\frac{6(3t-1)^3}{(2t+1)^3} \cdot \left(\frac{t+3}{2t+1}\right) \\ &= \frac{6(3t-1)^3 (t+3)}{(2t+1)^4}\\ \end{align}

Answer 2

it is $$f'(t)=4(3t-1)^3\cdot 3(2t+1)^{-3}+(3t-1)^4\cdot(-3)(2t+1)^{-4}\cdot 2$$ wich simplifies to $$6\,{\frac { \left( 3\,t-1 \right) ^{3} \left( t+3 \right) }{ \left( 2 \,t+1 \right) ^{4}}} $$ it is $$6(3t-1)^3\left(\frac{12}{(2t+1)^3}-\frac{6(3t-1)}{(2t+1)^4}\right)$$

Answer 3

Yes it is a correct application of

$$(t)=g(t)\cdot h(t)\implies f'(t)=g'(t)\cdot h(t)+g(t)\cdot h'(t)$$

that is

$$f'(t) = 12(3t-1)^3 \cdot (2t + 1)^{-3}-6(3t-1)^4 \cdot (2t + 1)^{-4}=$$

$$= \frac{12(3t-1)^3 \cdot (2t + 1)-6(3t-1)^4}{(2t + 1)^{4}}$$

$$= (3t-1)^3\frac{24t + 12-18t+6}{(2t + 1)^{4}}$$

$$= \frac{(3t-1)^3(6t +18)}{(2t + 1)^{4}}$$