Question

If \[x = \left( t + \frac{1}{t} \right)^a , y = a^{t + \frac{1}{t}} , \text{ find } \frac{dy}{dx}\] ?

Answer 1

\[\text { We have, x }= \left( t + \frac{1}{t} \right)^a\]

\[\Rightarrow \frac{dx}{dt} = \frac{d}{dt}\left[ \left( t + \frac{1}{t} \right)^a \right] \]
\[ \Rightarrow \frac{dx}{dt} = a \left( t + \frac{1}{t} \right)^{a - 1} \frac{d}{dt}\left( t + \frac{1}{t} \right)\]
\[ \Rightarrow \frac{dx}{dt} = a \left( t + \frac{1}{t} \right)^{a - 1} \left( 1 - \frac{1}{t^2} \right) . . . \left( i \right)\]
\[\text { and, } \]
\[ y = a^\left( t + \frac{1}{t} \right) \]

\[\Rightarrow \frac{dy}{dt} = \frac{d}{dt}\left[ a^\left( t + \frac{1}{t} \right) \right] \]
\[ \Rightarrow \frac{dy}{dt} = a^\left( t + \frac{1}{t} \right) \times \log a\frac{d}{dt}\left( t + \frac{t}{t} \right)\]
\[ \Rightarrow \frac{dy}{dt} = a^\left( t + \frac{1}{t} \right) \times \log a\left( 1 - \frac{1}{t^2} \right) . . . \left( ii \right)\]
\[\text { Dividing equation } \left( ii \right) by \left( i \right), \]
\[\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a^\left( t + \frac{1}{t} \right) \times \log a\left( 1 - \frac{1}{t^2} \right)}{a \left( t + \frac{1}{t} \right)^{a - 1} \left( 1 - \frac{1}{t^2} \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{a^\left( t + \frac{1}{t} \right) \times \log a}{a \left( t + \frac{1}{t} \right)^{a - 1}}\]