Question

How do I find the derivative of `lnroot3(4x-5)((3x+8)^2)`?

Answer 1

Holy Jesus Christ!

Explanation:

First, let's identify it: it's a product which two terms will depend on chain rule. Thus, we'll need both two rules.

• Product rule states that for `y=f(x)g(x)`, then `(dy)/(dx)=f'(x)g(x)+f(x)g'(x)`

• Chain rule states that `(dy)/(dx)=(dy)/(du)(du)/(dx)`

For the first term, let's rename `u=root(3)(v)` and `v=4x-5`.

For the second term, let's rename `w=3x+8`

There's no need to derivate it completely straightforward. You can do steps separately, but I do think that that might confuse things a bit.

`(dy)/(dx)=[(1/u)1/(3v^(2/3))(4x-5)](3x+8)^2+ln(root(3)(4x-5))2w(3)`

Substituting `u`:
`(dy)/(dx)=[(1/root(3)v)1/(3v^(2/3))(4x-5)](3x+8)^2+ln(root(3)(4x-5))2w(3)`

Substituting `v`:
`(dy)/(dx)=[(1/root(3)(4x-5))1/(3(4x-5)^(2/3))(4x-5)](3x+8)^2+ln(root(3)(4x-5))2w(3)`

Substituting `w`:
`(dy)/(dx)=[(1/root(3)(4x-5))1/(3(4x-5)^(2/3))(4x-5)](3x+8)^2+ln(root(3)(4x-5))(6(3x+8))`

`(dy)/(dx)=((4x-5)^cancel(2))/(3cancel((4x-5)))(3x+8)^2+ln(root(3)(4x-5))(6(3x+8))`

`(dy)/(dx)=(3x+8)[(((4x-5)(3x+8))/3)+6ln(root(3)(4x-5))]`

Feel free to develop it, but this is fine as a final answer!