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

1 Answer
Dec 26, 2016

Answer:

#v'(t)=120t^4-576t^3+888t^2-480t+96#

Explanation:

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

#v=(3t^2+1)(2t-4)^3#

By the product rule, #f(x)*g(x)=f'(x)g(x)+f(x)g'(x)#.

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

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

#f'(x)=6t#

The derivative of the second function, #g'(x)#, 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 #(2t-4)^3# as being composed of two functions, say #f(x)# and #g(x)#. We have #()^3# as one function and #(2t-4)# as its own function inside of the first. We can think of #( )^3# as being #f(x)# and #(2t-4)# as being #g(x)# in the above definition.

First, we take the derivative of the outermost function, #()^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(2t-4)^2#as the first half of our derivative. This is #f'(g(x))# in the above definition. Next, we multiply by the derivative of the inside function, #(2t-4)#, which is simply #2#.

#=>3(2t-4)^2*2#

#=>g'(x)=6(2t-4)^2#

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

#6t(2t-4)^3#

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

#=>6t(2t-4)^3+((3t^2+1)*6(2t-4)^2)#

#=>6t(2t-4)^3+6(3t^2+1)(2t-4)^2#

We can further simplify this answer. Factor out #(2t-4)^2#:

#=>(2t-4)^2[6t(2t-4)+6(3t^2+1)]#

Multiply out:

#=>(2t-4)^2(12t^2-24t+18t^2+6)#

#=>(2t-4)^2(30t^2-24t+6)#

Factor out #6#:

#=>6(2t-4)^2(5t^2-4t+1)#

Multiply out #(2t-4)^2#:

#=>6(4t^2-16t+16)(5t^2-4t+1)#

Multiply out:

#=>6(20t^4-16t^3+4t^2-80t^3+64t^2-16t+80t^2-64t+16)#

Factor out #4#

#=>24(5t^4color(red)(-4t^3)+color(blue)(t^2)color(red)(-20t^3)+color(blue)(16t^2)color(green)(-4t)+color(blue)(20t^2)color(green)(-16t)+4)#

Combine like terms:

#=>24(5t^4-24t^3+37t^2-20t+4)#

The final, simplified answer then is

#v'(t)=24(5t^4-24t^3+37t^2-20t+4)#

or

#v'(t)=120t^4-576t^3+888t^2-480t+96#