We are tasked with finding the derivative $\frac{dy}{dx}$ of the given function $y = \log_2(x^2)$ at $x = e^{-1}$.

Step-by-step solution:

1. Rewrite the function: The function is given as: $y = \log_2(x^2)$ Recall that $\log_a(b) = \frac{\ln(b)}{\ln(a)}$. Therefore: $y = \frac{\ln(x^2)}{\ln(2)}$

   Simplify further: $y = \frac{2\ln(x)}{\ln(2)}$

2. Differentiate $y$ with respect to $x$: Using the constant multiplier rule and the derivative of $\ln(x)$: $\frac{dy}{dx} = \frac{2}{\ln(2)} \cdot \frac{d}{dx}(\ln(x))$ Since $\frac{d}{dx}(\ln(x)) = \frac{1}{x}$, we get: $\frac{dy}{dx} = \frac{2}{\ln(2)} \cdot \frac{1}{x}$ Thus: $\frac{dy}{dx} = \frac{2}{x\ln(2)}$

3. Evaluate $\frac{dy}{dx}$ at $x = e^{-1}$: Substituting $x = e^{-1}$ into $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{2}{(e^{-1})\ln(2)}$ Simplify the denominator: $\frac{dy}{dx} = \frac{2}{\frac{1}{e} \cdot \ln(2)}$ $\frac{dy}{dx} = \frac{2e}{\ln(2)}$

Final Answer:

$\frac{dy}{dx} = \frac{2e}{\ln(2)} \quad \text{at } x = e^{-1}.$

Do you want further details or clarification?
Here are 5 related questions to explore further:

1. How do we derive the formula $\log_a(b) = \frac{\ln(b)}{\ln(a)}$?
2. Why is the derivative of $\ln(x)$ equal to $\frac{1}{x}$?
3. How do you compute the natural logarithm ($\ln$) of specific values like $\ln(2)$?
4. What are some other applications of logarithmic differentiation?
5. What happens if we use a different base logarithm, like $\log_{10}(x^2)$?

Tip: Always simplify logarithmic expressions before differentiating—it makes the process easier and reduces errors.