How do I find the derivative of lnroot3(4x-5)((3x+8)^2)ln34x5((3x+8)2)?

1 Answer
Dec 24, 2015

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)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!