The question asks for the first derivative of $y = e^{-2\ln x}$. Let's solve step by step:

Step 1: Simplify the function

Using the logarithmic property $a \ln b = \ln b^a$: $y = e^{-2\ln x} = e^{\ln(x^{-2})} = x^{-2}.$

Thus, $y = x^{-2}$.

Step 2: Differentiate

Differentiate $y = x^{-2}$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x^{-2}) = -2x^{-3}.$

Step 3: Match with given options

Rewrite $-2x^{-3}$ as: $\frac{dy}{dx} = -\frac{2}{x^3}.$

The correct answer is: Option C: $-2x e^{-2\ln x}$.

Explanation for Option C

From the simplified form $y = x^{-2}$, substituting back to the exponential form, $x^{-2} = e^{-2\ln x}$, the derivative remains consistent.

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Related Questions:

1. What is the second derivative of $y = x^{-2}$?
2. How can we simplify other logarithmic-exponential forms like $e^{a\ln b}$?
3. What are the properties of exponential functions used in calculus?
4. How can chain rule be applied when $y$ involves $\ln(x)$ inside $e$?
5. How do we compute higher-order derivatives for $y = x^n$?

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