2.2.206 AP Calculus Practice question derivative of a composite sine

[karush]

206 (day of year number)

If $f(x)=\sin{(\ln{(2x)})}$, then $f'(x)=$
(A) $\dfrac{\sin{(\ln{(2x)}}}{2x}$

(B) $\dfrac{\cos{(\ln{(2x)}}}{x}$

(C) $\dfrac{\cos{(\ln{(2x)}}}{2x}$

(D) $\cos{\left(\dfrac{1}{2x}\right)}$

Ok W|A returned (B) $\dfrac{\cos{(\ln{(2x)}}}{x}$
but I didn't understand why the $\ln(2x)$ was not changed?

these are being also posted in MeWe in the MathQuiz group

 

[Greg]

\(\displaystyle g(x)=\log(2x)=\log(2)+\log(x),\quad g'(x)=\frac1x\)

or, by the chain rule,

\(\displaystyle g'(x)=\frac{2}{2x}=\frac1x\)

 

[HallsofIvy]

By the chain rule, $$f(g(x))'= f'(g(x))g'(x)$$. Notice that the argument of f' is still g(x), not the derivative.