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!

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