If $x=\left(t+\frac{1}{t}\right)^{a}, y=a^{t+\frac{1}{t}}$, find $\frac{d y}{d x}$
We have, $x=\left(t+\frac{1}{t}\right)^{a}$
$\Rightarrow \frac{d x}{d t}=\frac{d}{d t}\left[\left(t+\frac{1}{t}\right)^{a}\right]$
$\Rightarrow \frac{d x}{d t}=a\left(t+\frac{1}{t}\right)^{a-1} \frac{d}{d t}\left(t+\frac{1}{t}\right)$
$\Rightarrow \frac{d x}{d t}=a\left(t+\frac{1}{t}\right)^{a-1}\left(1-\frac{1}{t^{2}}\right)$ ......(1)
and,
$y=a^{\left(t+\frac{1}{t}\right)}$
$\Rightarrow \frac{d y}{d t}=\frac{d}{d t}\left[a^{\left(t+\frac{1}{t}\right)}\right]$
$\Rightarrow \frac{d y}{d t}=a^{\left(t+\frac{1}{t}\right)} \times \log a \frac{d}{d t}\left(t+\frac{t}{t}\right)$
$\Rightarrow \frac{d y}{d t}=a^{\left(t+\frac{1}{t}\right)} \times \log a\left(1-\frac{1}{t^{2}}\right)$ ....(2)
Dividing equation (ii) by (i),
$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{a^{\left(t+\frac{1}{t}\right)} \times \log a\left(1-\frac{1}{t^{2}}\right)}{a\left(t+\frac{1}{t}\right)^{a-1}\left(1-\frac{1}{t^{2}}\right)}$
$\Rightarrow \frac{d y}{d x}=\frac{a^{\left(t+\frac{1}{t}\right)} \times \log a}{a\left(t+\frac{1}{t}\right)^{a-1}}$
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