18. [0/1 Points] DETAILS MY NOTES Use logarithmic differentiation or an alternative method to find the derivative of the functio \[ y=(\sin 7 x)^{\ln x} \] \( y^{\prime}=\square \)

Let \( y = (\sin 7x)^{\ln x} \). To differentiate using logarithmic differentiation, we first take the natural logarithm of both sides: \[ \ln y = \ln\left((\sin 7x)^{\ln x}\right) \] Using the logarithm property \( \ln(a^b)= b \ln a \), we have \[ \ln y = \ln x \cdot \ln(\sin 7x) \] Now, differentiate both sides with respect to \( x \). On the left side, use the chain rule: \[ \frac{d}{dx}(\ln y) = \frac{y'}{y} \] For the right side, differentiate the product \( \ln x \cdot \ln(\sin 7x) \) using the product rule: \[ \frac{d}{dx}\left[\ln x \cdot \ln(\sin 7x)\right] = \left(\frac{1}{x}\right) \ln(\sin 7x) + \ln x \cdot \frac{d}{dx}\left[\ln(\sin 7x)\right] \] Differentiate \( \ln(\sin 7x) \) using the chain rule: \[ \frac{d}{dx}\left[\ln(\sin 7x)\right] = \frac{1}{\sin 7x} \cdot (\cos 7x \cdot 7) = \frac{7\cos 7x}{\sin 7x} \] Thus, the derivative of the right side is: \[ \frac{1}{x} \ln(\sin 7x) + \ln x \cdot \frac{7\cos 7x}{\sin 7x} \] So we have: \[ \frac{y'}{y} = \frac{\ln(\sin 7x)}{x} + \frac{7\ln x\, \cos 7x}{\sin 7x} \] Finally, multiply both sides by \( y \) to solve for \( y' \): \[ y' = y \left( \frac{\ln(\sin 7x)}{x} + \frac{7\ln x\, \cos 7x}{\sin 7x} \right) \] Recall that \( y = (\sin 7x)^{\ln x} \), hence: \[ y' = (\sin 7x)^{\ln x} \left( \frac{\ln(\sin 7x)}{x} + \frac{7\ln x\, \cos 7x}{\sin 7x} \right) \]