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

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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
$$
y' = 7x^6 \sin(x) + x^7 \cos(x).
$$
The derivative of $$ y = x^7 \sin(x) $$ is $$ y' = 7x^6 \sin(x) + x^7 \cos(x) $$.

Use logarithmic differentlation to find the derivative of the function.

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