How do you differentiate #y = ln root3(6x + 7)#?

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Answer 1 · 2017-01-07T15:05:47

#dy/dx = 2/(6x + 7)#

Rewrite using laws of logarithms:

#y = ln(6x + 7)^(1/3)#

#y = 1/3ln(6x + 7)#

#3y = ln(6x + 7)#

Now, let #y= lnu# and #u = 6x + 7#. Then #dy/(du) = 1/u# and #(du)/dx = 6#.

#dy/dx = dy/(du) * (du)/dx#

#dy/dx = 1/u * 6#

#dy/dx = 1/(6x + 7) * 6#

#dy/dx = 6/(6x + 7)#

Differentiate the left hand side now implicitly.

#d/dx(3y) = 3(dy/dx)#

Put this together:

#3dy/dx = 6/(6x + 7)#

#dy/dx = 6/(3(6x + 7))#

#dy/dx = 2/(6x + 7)#

Hopefully this helps!