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Let $u = a^{t+\frac{1}{t}}$ and $v = (t+\frac{1}{t})^a$. We need to find $\frac{du}{dv}$.
We have $u = a^{t+\frac{1}{t}}$. Taking the natural logarithm of both sides, we get $\ln u = (t+\frac{1}{t}) \ln a$. Differentiating both sides with respect to $t$, we have $$ \frac{1}{u} \frac{du}{dt} = \ln a \cdot (1 - \frac{1}{t^2}) $$ $$ \frac{du}{dt} = u \ln a \cdot (1 - \frac{1}{t^2}) = a^{t+\frac{1}{t}} \ln a \cdot (1 - \frac{1}{t^2}) $$
We have $v = (t+\frac{1}{t})^a$. Taking the natural logarithm of both sides, we get $\ln v = a \ln (t+\frac{1}{t})$. Differentiating both sides with respect to $t$, we have $$ \frac{1}{v} \frac{dv}{dt} = a \cdot \frac{1}{t+\frac{1}{t}} \cdot (1 - \frac{1}{t^2}) $$ $$ \frac{dv}{dt} = v \cdot a \cdot \frac{1}{t+\frac{1}{t}} \cdot (1 - \frac{1}{t^2}) = (t+\frac{1}{t})^a \cdot a \cdot \frac{1}{t+\frac{1}{t}} \cdot (1 - \frac{1}{t^2}) = a (t+\frac{1}{t})^{a-1} (1 - \frac{1}{t^2}) $$
We need to find $\frac{du}{dv} = \frac{du/dt}{dv/dt}$. $$ \frac{du}{dv} = \frac{a^{t+\frac{1}{t}} \ln a \cdot (1 - \frac{1}{t^2})}{a (t+\frac{1}{t})^{a-1} (1 - \frac{1}{t^2})} = \frac{a^{t+\frac{1}{t}} \ln a}{a (t+\frac{1}{t})^{a-1}} $$ $$ \frac{du}{dv} = \frac{a^{t+\frac{1}{t}-1} \ln a}{(t+\frac{1}{t})^{a-1}} $$
Final Answer: $\frac{a^{t+\frac{1}{t}-1} \ln a}{(t+\frac{1}{t})^{a-1}}$
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