Use logarithmic differentlation to find the derivative of the function. \[ y=x^{7} \sin (x) \] \( y^{\prime}(x)=1 \)

Let \[ y = x^7 \sin(x). \] Taking the natural logarithm of both sides, we have \[ \ln y = \ln(x^7 \sin(x)). \] Using the logarithm property \(\ln(ab) = \ln a + \ln b\) and \(\ln(x^7)=7\ln x\), this becomes \[ \ln y = 7 \ln x + \ln \sin(x). \] Next, differentiate both sides with respect to \(x\). Using the chain rule on the left side, we get \[ \frac{y'}{y} = \frac{d}{dx}\left(7 \ln x\right) + \frac{d}{dx}\left(\ln \sin(x)\right). \] The derivative of \(7\ln x\) is \[ \frac{d}{dx}\left(7\ln x\right) = \frac{7}{x}. \] For the derivative of \(\ln \sin(x)\), we use the chain rule: \[ \frac{d}{dx}\left(\ln \sin(x)\right) = \frac{1}{\sin(x)} \cos(x) = \cot(x). \] Thus, we have \[ \frac{y'}{y} = \frac{7}{x} + \cot(x). \] To solve for \(y'\), multiply both sides by \(y\): \[ y' = y \left(\frac{7}{x} + \cot(x)\right). \] Since \(y = x^7 \sin(x)\), substitute it back: \[ y' = x^7 \sin(x) \left(\frac{7}{x} + \cot(x)\right). \] Recognize that \(\cot(x) = \frac{\cos(x)}{\sin(x)}\). Therefore, distribute: \[ y' = x^7 \sin(x) \left(\frac{7}{x}\right) + x^7 \sin(x) \left(\frac{\cos(x)}{\sin(x)}\right). \] Simplify both terms: \[ y' = 7x^6 \sin(x) + x^7 \cos(x). \] Thus, the derivative is \[ y' = 7x^6 \sin(x) + x^7 \cos(x). \]