`y=(ln\ 2x)/(e^(2x)+2`

We let u = ln 2x and v = e2x + 2, and we'll use the derivative of a quotient formula,

`(dy)/(dx)=(v(du)/(dx)-u(dv)/(dx))/(v^2)`

We'll need to find the derivative of both u and v before using the formula.

Now, using the logarithm laws, we have:

u = ln 2x

= ln 2 + ln x.

So simply,

`(du)/(dx)=1/x`

And for v = e2x + 2 we have:

`(dv)/(dx)=2e^(2x)`

So applying the quotient formula, we obtain:

`(dy)/(dx)=((e^(2x)+2)(d(ln\ 2x))/(dx)-ln\ 2x(d(e^(2x)+2))/(dx))/((e^(2x)+2)^2`

Using the derivatives we just found for u and v gives:

`(dy)/(dx)=((e^(2x)+2)(1/x)-ln\ 2x(2e^(2x)))/((e^(2x)+2)^2`

We tidy this up by multiplying top and bottom by x. Also, it's best to write the 2e2x in front of the logarithm expression to reduce confusion. (It's not part of the log expression.)

So our derivative is:

`(dy)/(dx)=((e^(2x)+2)-2e^(2x)(x)(ln\ 2x))/(x(e^(2x)+2)^2)`

`(dy)/(dx)=((e^(2x)+2)-2x\ e^(2x)(ln\ 2x))/(x(e^(2x)+2)^2)`

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