Derivative of $$$\ln\left(x\right) \sin{\left(9 x \right)}$$$

Let $$$H{\left(x \right)} = \ln\left(x\right) \sin{\left(9 x \right)}$$$.

Take the logarithm of both sides: $$$\ln\left(H{\left(x \right)}\right) = \ln\left(\ln\left(x\right) \sin{\left(9 x \right)}\right)$$$.

Rewrite the RHS using the properties of logarithms: $$$\ln\left(H{\left(x \right)}\right) = \ln\left(\ln\left(x\right)\right) + \ln\left(\sin{\left(9 x \right)}\right)$$$.

Differentiate separately both sides of the equation: $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right) + \ln\left(\sin{\left(9 x \right)}\right)\right)$$$.

Differentiate the LHS of the equation.

The function $$$\ln\left(H{\left(x \right)}\right)$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \ln\left(u\right)$$$ and $$$g{\left(x \right)} = H{\left(x \right)}$$$.

Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

$${\color{red}\left(\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(H{\left(x \right)}\right)\right)}$$

The derivative of the natural logarithm is $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(H{\left(x \right)}\right) = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(H{\left(x \right)}\right)$$

Return to the old variable:

$$\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{{\color{red}\left(u\right)}} = \frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{{\color{red}\left(H{\left(x \right)}\right)}}$$

Thus, $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}}$$$.

Differentiate the RHS of the equation.

The derivative of a sum/difference is the sum/difference of derivatives:

$${\color{red}\left(\frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right) + \ln\left(\sin{\left(9 x \right)}\right)\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right)\right) + \frac{d}{dx} \left(\ln\left(\sin{\left(9 x \right)}\right)\right)\right)}$$

The function $$$\ln\left(\sin{\left(9 x \right)}\right)$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \ln\left(u\right)$$$ and $$$g{\left(x \right)} = \sin{\left(9 x \right)}$$$.

Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

$${\color{red}\left(\frac{d}{dx} \left(\ln\left(\sin{\left(9 x \right)}\right)\right)\right)} + \frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right)\right) = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(\sin{\left(9 x \right)}\right)\right)} + \frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right)\right)$$

The derivative of the natural logarithm is $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(\sin{\left(9 x \right)}\right) + \frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right)\right) = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(\sin{\left(9 x \right)}\right) + \frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right)\right)$$

Return to the old variable:

$$\frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right)\right) + \frac{\frac{d}{dx} \left(\sin{\left(9 x \right)}\right)}{{\color{red}\left(u\right)}} = \frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right)\right) + \frac{\frac{d}{dx} \left(\sin{\left(9 x \right)}\right)}{{\color{red}\left(\sin{\left(9 x \right)}\right)}}$$

The function $$$\sin{\left(9 x \right)}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ and $$$g{\left(x \right)} = 9 x$$$.

Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

$$\frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right)\right) + \frac{{\color{red}\left(\frac{d}{dx} \left(\sin{\left(9 x \right)}\right)\right)}}{\sin{\left(9 x \right)}} = \frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right)\right) + \frac{{\color{red}\left(\frac{d}{du} \left(\sin{\left(u \right)}\right) \frac{d}{dx} \left(9 x\right)\right)}}{\sin{\left(9 x \right)}}$$

The derivative of the sine is $$$\frac{d}{du} \left(\sin{\left(u \right)}\right) = \cos{\left(u \right)}$$$:

$$\frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right)\right) + \frac{{\color{red}\left(\frac{d}{du} \left(\sin{\left(u \right)}\right)\right)} \frac{d}{dx} \left(9 x\right)}{\sin{\left(9 x \right)}} = \frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right)\right) + \frac{{\color{red}\left(\cos{\left(u \right)}\right)} \frac{d}{dx} \left(9 x\right)}{\sin{\left(9 x \right)}}$$

Return to the old variable:

$$\frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right)\right) + \frac{\cos{\left({\color{red}\left(u\right)} \right)} \frac{d}{dx} \left(9 x\right)}{\sin{\left(9 x \right)}} = \frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right)\right) + \frac{\cos{\left({\color{red}\left(9 x\right)} \right)} \frac{d}{dx} \left(9 x\right)}{\sin{\left(9 x \right)}}$$

The function $$$\ln\left(\ln\left(x\right)\right)$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \ln\left(u\right)$$$ and $$$g{\left(x \right)} = \ln\left(x\right)$$$.

Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

$${\color{red}\left(\frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right)\right)\right)} + \frac{\cos{\left(9 x \right)} \frac{d}{dx} \left(9 x\right)}{\sin{\left(9 x \right)}} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(\ln\left(x\right)\right)\right)} + \frac{\cos{\left(9 x \right)} \frac{d}{dx} \left(9 x\right)}{\sin{\left(9 x \right)}}$$

The derivative of the natural logarithm is $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(\ln\left(x\right)\right) + \frac{\cos{\left(9 x \right)} \frac{d}{dx} \left(9 x\right)}{\sin{\left(9 x \right)}} = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(\ln\left(x\right)\right) + \frac{\cos{\left(9 x \right)} \frac{d}{dx} \left(9 x\right)}{\sin{\left(9 x \right)}}$$

Return to the old variable:

$$\frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{{\color{red}\left(u\right)}} + \frac{\cos{\left(9 x \right)} \frac{d}{dx} \left(9 x\right)}{\sin{\left(9 x \right)}} = \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{{\color{red}\left(\ln\left(x\right)\right)}} + \frac{\cos{\left(9 x \right)} \frac{d}{dx} \left(9 x\right)}{\sin{\left(9 x \right)}}$$

The derivative of the natural logarithm is $$$\frac{d}{dx} \left(\ln\left(x\right)\right) = \frac{1}{x}$$$:

$$\frac{\cos{\left(9 x \right)} \frac{d}{dx} \left(9 x\right)}{\sin{\left(9 x \right)}} + \frac{{\color{red}\left(\frac{d}{dx} \left(\ln\left(x\right)\right)\right)}}{\ln\left(x\right)} = \frac{\cos{\left(9 x \right)} \frac{d}{dx} \left(9 x\right)}{\sin{\left(9 x \right)}} + \frac{{\color{red}\left(\frac{1}{x}\right)}}{\ln\left(x\right)}$$

Apply the constant multiple rule $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ with $$$c = 9$$$ and $$$f{\left(x \right)} = x$$$:

$$\frac{\cos{\left(9 x \right)} {\color{red}\left(\frac{d}{dx} \left(9 x\right)\right)}}{\sin{\left(9 x \right)}} + \frac{1}{x \ln\left(x\right)} = \frac{\cos{\left(9 x \right)} {\color{red}\left(9 \frac{d}{dx} \left(x\right)\right)}}{\sin{\left(9 x \right)}} + \frac{1}{x \ln\left(x\right)}$$

Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:

$$\frac{9 \cos{\left(9 x \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{\sin{\left(9 x \right)}} + \frac{1}{x \ln\left(x\right)} = \frac{9 \cos{\left(9 x \right)} {\color{red}\left(1\right)}}{\sin{\left(9 x \right)}} + \frac{1}{x \ln\left(x\right)}$$

Simplify:

$$\frac{9 \cos{\left(9 x \right)}}{\sin{\left(9 x \right)}} + \frac{1}{x \ln\left(x\right)} = \frac{9}{\tan{\left(9 x \right)}} + \frac{1}{x \ln\left(x\right)}$$

Thus, $$$\frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right) + \ln\left(\sin{\left(9 x \right)}\right)\right) = \frac{9}{\tan{\left(9 x \right)}} + \frac{1}{x \ln\left(x\right)}$$$.

Hence, $$$\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}} = \frac{9}{\tan{\left(9 x \right)}} + \frac{1}{x \ln\left(x\right)}$$$.

Therefore, $$$\frac{d}{dx} \left(H{\left(x \right)}\right) = \left(\frac{9}{\tan{\left(9 x \right)}} + \frac{1}{x \ln\left(x\right)}\right) H{\left(x \right)} = 9 \ln\left(x\right) \cos{\left(9 x \right)} + \frac{\sin{\left(9 x \right)}}{x}$$$.