Derivative of $$$\ln\left(\sin{\left(x \right)}\right)$$$
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Solution
The function $$$\ln\left(\sin{\left(x \right)}\right)$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \ln\left(u\right)$$$ and $$$g{\left(x \right)} = \sin{\left(x \right)}$$$.
Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(\ln\left(\sin{\left(x \right)}\right)\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(\sin{\left(x \right)}\right)\right)}$$The derivative of the natural logarithm is $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:
$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(\sin{\left(x \right)}\right) = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(\sin{\left(x \right)}\right)$$Return to the old variable:
$$\frac{\frac{d}{dx} \left(\sin{\left(x \right)}\right)}{{\color{red}\left(u\right)}} = \frac{\frac{d}{dx} \left(\sin{\left(x \right)}\right)}{{\color{red}\left(\sin{\left(x \right)}\right)}}$$The derivative of the sine is $$$\frac{d}{dx} \left(\sin{\left(x \right)}\right) = \cos{\left(x \right)}$$$:
$$\frac{{\color{red}\left(\frac{d}{dx} \left(\sin{\left(x \right)}\right)\right)}}{\sin{\left(x \right)}} = \frac{{\color{red}\left(\cos{\left(x \right)}\right)}}{\sin{\left(x \right)}}$$Simplify:
$$\frac{\cos{\left(x \right)}}{\sin{\left(x \right)}} = \frac{1}{\tan{\left(x \right)}}$$Thus, $$$\frac{d}{dx} \left(\ln\left(\sin{\left(x \right)}\right)\right) = \frac{1}{\tan{\left(x \right)}}$$$.
Answer
$$$\frac{d}{dx} \left(\ln\left(\sin{\left(x \right)}\right)\right) = \frac{1}{\tan{\left(x \right)}}$$$A