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Question:

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

Solution:

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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