Find $$ \frac{d y}{d x} $$ if: $$ y=x \ln \left(\frac{1}{x}\right) $$

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Let
$$
y = x \ln\left(\frac{1}{x}\right).
$$
First, recall that
$$
\ln\left(\frac{1}{x}\right) = -\ln x.
$$
Thus, the function becomes
$$
y = x \left(-\ln x\right) = -x \ln x.
$$

Next, we differentiate using the product rule. The product rule states that if
$$
f(x) = u(x)v(x),
$$
then
$$
f'(x) = u'(x)v(x) + u(x)v'(x).
$$

Here, take
$$
u(x) = -x \quad \text{and} \quad v(x) = \ln x.
$$
Differentiate each component:
$$
u'(x) = -1, \quad v'(x) = \frac{1}{x}.
$$

Now, apply the product rule:
$$
\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) = (-1)(\ln x) + (-x)\left(\frac{1}{x}\right).
$$
Simplify the expression:
$$
\frac{dy}{dx} = -\ln x - 1.
$$

Thus, the derivative is:
$$
\frac{dy}{dx} = -\left(\ln x + 1\right).
$$
The derivative $$ \frac{dy}{dx} $$ is $$ -\ln x - 1 $$.

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