If \( x = e^{t + \frac{1}{t}} \) and \( y = e^{t - \frac{1}{t}} \), then find \( \frac{dy}{dx} \) at \( t = -2 \).
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Instead of dealing with fractions throughout, simplifying the quotients of exponential functions early using laws of exponents (\( \frac{e^A}{e^B} = e^{A-B} \)) keeps the algebra clean and reduces substitution errors at the end.