At ferentiate with respect to $$ x $$ in the retation $$ \left(\frac{1}{x}+\log y\right) $$

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To differentiate the expression $$ \left(\frac{1}{x} + \log y\right) $$ with respect to $$ x $$, we will apply the rules of differentiation, including the product rule and the chain rule where necessary.

1. **Differentiate $$ \frac{1}{x} $$**:
   - The derivative of $$ \frac{1}{x} $$ is $$ -\frac{1}{x^2} $$.

2. **Differentiate $$ \log y $$**:
   - Since $$ y $$ is a function of $$ x $$, we will use the chain rule. The derivative of $$ \log y $$ with respect to $$ x $$ is $$ \frac{1}{y} \cdot \frac{dy}{dx} $$.

Now, we can combine these results:

$$
\frac{d}{dx}\left(\frac{1}{x} + \log y\right) = -\frac{1}{x^2} + \frac{1}{y} \cdot \frac{dy}{dx}
$$

Thus, the derivative of the expression $$ \left(\frac{1}{x} + \log y\right) $$ with respect to $$ x $$ is:

$$
-\frac{1}{x^2} + \frac{1}{y} \cdot \frac{dy}{dx}
$$
The derivative of $$ \left(\frac{1}{x} + \log y\right) $$ with respect to $$ x $$ is $$ -\frac{1}{x^2} + \frac{1}{y} \cdot \frac{dy}{dx} $$.

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