How do you find the derivative of #v=(3t^2+1)(2t-4)^3#?
1 Answer
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,
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)=6t#
The derivative of the second function,
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[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(2t-4)^2(5t^2-4t+1)#
Multiply out
#=>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
#=>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#
