First, we use the following log laws to simplify our logarithm expression:

log ab = log a + log b

and

log an = n log a

So we can write our question as:

`y=ln[(sin\ 2x)(sqrt(x^2+1))]`

`=ln(sin\ 2x)+ln(sqrt(x^2+1))`

`=ln(sin\ 2x)+ln(x^2+1)^(1/2)`

`=ln(sin\ 2x)+1/2ln(x^2+1)`

Next, we use the following rule (twice) to differentiate the two log terms:

`(dy)/(dx)=(u’)/u`

For the first term,

u = sin 2x

u' = 2 cos 2x

For the second term, we put

u = x2 + 1,

giving

u' = 2x

So our final answer is:

`(dy)/(dx)=(2\ cos\ 2x)/(sin\ 2x)+1/2 (2x)/(x^2+1)`

`=2\ cot\ 2x+x/(x^2+1)`