Solve this

$a s x=\left(t+\frac{1}{t}\right)^{a}$

Differentiating it with respect to $t$ using chain rule,

$\frac{d x}{d t}=\frac{d}{d t}\left(\left(t+\frac{1}{t}\right)^{a}\right)$

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

$\frac{d x}{d t}=a\left(\left(t+\frac{1}{t}\right)^{a-1}\right)\left(1-\frac{1}{t^{2}}\right) \ldots . .$(1)

And, $\mathrm{y}=\mathrm{a}^{\left(t+\frac{1}{t}\right)}$

Differentiating it with respect to t using chain rule,

$\frac{d y}{d t}=\frac{d}{d t}\left(a^{\left(t+\frac{1}{t}\right)}\right)$

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

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

Dividing equation (2) by (1),

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

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