Question

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

Answer 1

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 :)