How do you find the derivative of v=(3t^2+1)(2t-4)^3?

Dec 26, 2016

$v ' \left(t\right) = 120 {t}^{4} - 576 {t}^{3} + 888 {t}^{2} - 480 t + 96$

Explanation:

You will need to use both the product and chain rule to derive this function.

$v = \left(3 {t}^{2} + 1\right) {\left(2 t - 4\right)}^{3}$

By the product rule, $f \left(x\right) \cdot g \left(x\right) = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$.

In this case, $f \left(x\right) = 3 {t}^{2} + 1$ and $g \left(x\right) = {\left(2 t - 4\right)}^{3}$. You may find it helpful to determine $f ' \left(x\right)$ and $g ' \left(x\right)$ first.

$f ' \left(x\right)$ will be a fairly basic derivative. First we can derive $3 {t}^{2}$; bring down the power and reduce the power by one. The derivative of any constant is $0$, so we have:

$f ' \left(x\right) = 6 t$

The derivative of the second function, $g ' \left(x\right)$, will involve the chain rule, which is given by f@g(x))'=f'(g(x))g'(x). This may look a bit confusing, but in reality it's not so bad once you get the hang of it.

We can think of ${\left(2 t - 4\right)}^{3}$ as being composed of two functions, say $f \left(x\right)$ and $g \left(x\right)$. We have ${\left(\right)}^{3}$ as one function and $\left(2 t - 4\right)$ as its own function inside of the first. We can think of ${\left(\right)}^{3}$ as being $f \left(x\right)$ and $\left(2 t - 4\right)$ as being $g \left(x\right)$ in the above definition.

First, we take the derivative of the outermost function, ${\left(\right)}^{3}$, and leave the inner function as is. The derivative of some term raised to a power is found by multiplying the term by the original power and reducing the power by one. So we have $3 {\left(2 t - 4\right)}^{2}$as the first half of our derivative. This is $f ' \left(g \left(x\right)\right)$ in the above definition. Next, we multiply by the derivative of the inside function, $\left(2 t - 4\right)$, which is simply $2$.

$\implies 3 {\left(2 t - 4\right)}^{2} \cdot 2$

$\implies g ' \left(x\right) = 6 {\left(2 t - 4\right)}^{2}$

Now we can use the product rule. We multiply the derivative of the first function by the second function, which we leave alone:

$6 t {\left(2 t - 4\right)}^{3}$

Then we add to this the product of the derivative of the second function and the original first function:

$\implies 6 t {\left(2 t - 4\right)}^{3} + \left(\left(3 {t}^{2} + 1\right) \cdot 6 {\left(2 t - 4\right)}^{2}\right)$

$\implies 6 t {\left(2 t - 4\right)}^{3} + 6 \left(3 {t}^{2} + 1\right) {\left(2 t - 4\right)}^{2}$

We can further simplify this answer. Factor out ${\left(2 t - 4\right)}^{2}$:

$\implies {\left(2 t - 4\right)}^{2} \left[6 t \left(2 t - 4\right) + 6 \left(3 {t}^{2} + 1\right)\right]$

Multiply out:

$\implies {\left(2 t - 4\right)}^{2} \left(12 {t}^{2} - 24 t + 18 {t}^{2} + 6\right)$

$\implies {\left(2 t - 4\right)}^{2} \left(30 {t}^{2} - 24 t + 6\right)$

Factor out $6$:

$\implies 6 {\left(2 t - 4\right)}^{2} \left(5 {t}^{2} - 4 t + 1\right)$

Multiply out ${\left(2 t - 4\right)}^{2}$:

$\implies 6 \left(4 {t}^{2} - 16 t + 16\right) \left(5 {t}^{2} - 4 t + 1\right)$

Multiply out:

$\implies 6 \left(20 {t}^{4} - 16 {t}^{3} + 4 {t}^{2} - 80 {t}^{3} + 64 {t}^{2} - 16 t + 80 {t}^{2} - 64 t + 16\right)$

Factor out $4$

$\implies 24 \left(5 {t}^{4} \textcolor{red}{- 4 {t}^{3}} + \textcolor{b l u e}{{t}^{2}} \textcolor{red}{- 20 {t}^{3}} + \textcolor{b l u e}{16 {t}^{2}} \textcolor{g r e e n}{- 4 t} + \textcolor{b l u e}{20 {t}^{2}} \textcolor{g r e e n}{- 16 t} + 4\right)$

Combine like terms:

$\implies 24 \left(5 {t}^{4} - 24 {t}^{3} + 37 {t}^{2} - 20 t + 4\right)$

The final, simplified answer then is

$v ' \left(t\right) = 24 \left(5 {t}^{4} - 24 {t}^{3} + 37 {t}^{2} - 20 t + 4\right)$

or

$v ' \left(t\right) = 120 {t}^{4} - 576 {t}^{3} + 888 {t}^{2} - 480 t + 96$