How do you find the derivative of # [((3x-1)^4)((2x+1)^-3)]# using the chain rule?

1 Answer
Jan 17, 2018

Answer:

You going to have to use both the product and the chain rule.

#(dy)/(dx)=((3x-1)^3)/((2x+1)^(3))[12-6((3x-1))/((2x+1)]]#

Explanation:

Just to refresh your memory;
Product Rule:
#y=f(x)xxg(x)#

#(dy)/(dx)=f'(x)xxg(x)+f(x)xxg'(x)#

Chain Rule:
#y=f(g(x))#

#(dy)/(dx)=f'(g(x))xxg'(x)#

#y=[((3x−1)^4)((2x+1)^(−3))]#

First we will break it into the front #((3x−1)^4)# and back #((2x+1)^(−3))# sections and differentiate them individually.

#color(red)(first Section)#
#[d(3x−1)^4]/(dx)=4(3x-1)^3xx3=12(3x-1)^3#

#color(red)(Second Section)#

#[d(2x+1)^(-3)]/(dx)=-3(2x+1)^(-4)xx2=-6(2x+1)^-4#

Finally, using the product rule we combine these two sections to form the total derivative.

#(dy)/(dx)=(12(3x-1)^3)((2x+1)^(-3))+((3x−1)^4)(-6(2x+1)^-4)#

The easiest simplification to do is to move the negative sign out the front of the second section so it become a subtraction rather than an addition.

#(dy)/(dx)=(12(3x-1)^3)((2x+1)^(-3))-((3x−1)^4)(6(2x+1)^-4)#

You could then move the brackets with negative exponents into the denominator.

#(dy)/(dx)=(12(3x-1)^3)/((2x+1)^(3))-(6(3x−1)^4)/((2x+1)^4)#

Finally, you could factorise by #((3x-1)^3)/((2x+1)^(3)#

#(dy)/(dx)=((3x-1)^3)/((2x+1)^(3))[12-6((3x-1))/((2x+1)]]#

I hope that helps :)