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

The calculator will find the derivative of $\ln\left(x\right) \sin{\left(9 x \right)}$ using the logarithmic differentiation, with steps shown.

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Find $\frac{d}{dx} \left(\ln\left(x\right) \sin{\left(9 x \right)}\right)$.

### Solution

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 $\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$:

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

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

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 $\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$:

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

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

The function $\ln\left(\ln\left(x\right)\right)$ 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 $\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$:

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

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

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

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

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

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

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

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

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

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

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

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

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