To find $\frac{dy}{dx}$ for $y = \log_e \left(\frac{1}{x}\right)$, we will use the chain rule. Step 1: Define a substitution for the inner function. Let $u = \frac{1}{x}$. Then $y = \log_e u$. Step 2: Differentiate $y$ with respect to $u$. $$ \frac{dy}{du} = \frac{d}{du}(\log_e u) = \frac{1}{u} $$ Step 3: Differentiate $u$ with respect to $x$. We can write $u = x^{-1}$. $$ \frac{du}{dx} = \frac{d}{dx}(x^{-1}) = -1 \cdot x^{-1-1} = -x^{-2} = -\frac{1}{x^2} $$ Step 4: Apply the chain rule formula $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$. $$ \frac{dy}{dx} = \left(\frac{1}{u}\right) \times \left(-\frac{1}{x^2}\right) $$ Step 5: Substitute $u = \frac{1}{x}$ back into the expression for $\frac{dy}{dx}$. $$ \frac{dy}{dx} = \left(\frac{1}{\frac{1}{x}}\right) \times \left(-\frac{1}{x^2}\right) $$ $$ \frac{dy}{dx} = (x) \times \left(-\frac{1}{x^2}\right) $$ $$ \frac{dy}{dx} = -\frac{x}{x^2} $$ $$ \frac{dy}{dx} = -\frac{1}{x} $$ Comparing this result with the given options, the correct option is A. The final answer is $\boxed{\text{A. }-\frac{1}{x}}$. Send me the next one 📸