Question

Can someone explain :
Derivative
y = sqrt(ln(6x-1))

Answer 1

chain rule for this one

Answer 2

\[(f\circ g)'(x)=f'(g(x))\times g'(x)\] with
\[f(x)=\sqrt{x}, f'(x)=\frac{1}{2\sqrt{x}},g(x)=\ln(6x-1), g'(x)=\frac{6}{6x-1}\]

Answer 3

more simply put, the derivative of the square root of something is one over two square root of something, times the derivative of something

Answer 4

Which is considered the outside, and the inside? \[1/2(1/u)\]?

or is everything inside the square root considered the inside so \[1/2(u) \] Where u = ln(6x-1)
or
\[1/2(1/u)\]
where u = 6x-1

Answer 5

I responded without refreshing... alright so

I would sub in g(x)
\[\frac{ 1 }{ 2\sqrt{\ln(6x-1)} }\]
Then multiple g'(x) by the f'(g(x) above?

Answer 6

No updates, bad load times, can't respond...
If this ever goes through this is what I figured

\[\frac{ 1 }{ 2\sqrt{\ln(6x-1)} }*\frac{ 6 }{ 6x-1 }\]

\[\frac{ 6 }{ 12x-2\sqrt{\ln(6x-1}}\]
Final :
\[\frac{ 3 }{ 12x-\sqrt{\ln(6x-1)} }\]