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

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!