s(x) = ln (sin 2x) + sin (ln (2x))

[Sophie]

The solution seems too simple have I missed something silly?

Find the derivative

\(\displaystyle \L\\\begin{array}{l}
s(x) = \ln (\sin 2x) + \sin (\ln (2x)) \\

s'(x) = \frac{{2\cos 2x}}{{\sin 2x}} + \frac{{2\cos (\ln (2x)}}{{2x}} \\
s'(x) = 2\cot 2x + \frac{{\cos (\ln (2x)}}{x} \\
\end{array}\)


Thanks Sophie

 

[mark07]

You got it right.

 

[Sophie]

thanks