For a positive constant $a$ find $\frac{dy}{dx}$, where $y = a^{t + \frac{1}{t}}, \text{ and } x = \left( t + \frac{1}{t} \right)^a$.

$\frac{a^{t + \frac{1}{t}} \ln a}{a (t + \frac{1}{t})^{a-1}}$

The correct answer is Option (1) → $\frac{a^{t + \frac{1}{t}} \ln a}{a (t + \frac{1}{t})^{a-1}}$ ##

Observe that both $y$ and $x$ are defined for all real $t \neq 0$. Clearly

$ \frac{dy}{dt} = \frac{d}{dt} \left( a^{t + \frac{1}{t}} \right) = a^{t + \frac{1}{t}} \frac{d}{dt} \left( t + \frac{1}{t} \right) \cdot \log a$

$= a^{t + \frac{1}{t}} \left( 1 - \frac{1}{t^2} \right) \log a$

Similarly $ \frac{dx}{dt} = a \left[ t + \frac{1}{t} \right]^{a-1} \cdot \frac{d}{dt} \left( t + \frac{1}{t} \right)$

$= a \left[ t + \frac{1}{t} \right]^{a-1} \cdot \left( 1 - \frac{1}{t^2} \right)$

$\frac{dx}{dt} \neq 0$ only if $t \neq \pm 1$. Thus for $t \neq \pm 1$,

$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a^{t + \frac{1}{t}} \left( 1 - \frac{1}{t^2} \right) \log a}{a \left[ t + \frac{1}{t} \right]^{a-1} \cdot \left( 1 - \frac{1}{t^2} \right)}$

$= \frac{a^{t + \frac{1}{t}} \log a}{a \left( t + \frac{1}{t} \right)^{a-1}}$