For a positive constant a,find frac{dy}{dx},w | vedclass

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Question

(A) Given $y = a^{t+\frac{1}{t}}$ and $x = \left(t+\frac{1}{t}\right)^{a}$.First,differentiate $y$ with respect to $t$ using the chain rule:$\frac{dy}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $y = a^{t+\frac{1}{t}}$ and $x = \left(t+\frac{1}{t}ight)^{a}$.First,differentiate $y$ with respect to $t$ using the chain rule:$\frac{dy}{dt} = \frac{d}{dt}\left(a^{t+\frac{1}{t}}ight) = a^{t+\frac{1}{t}} \cdot \log a \cdot \frac{d}{dt}\left(t+\frac{1}{t}ight) = a^{t+\frac{1}{t}} \cdot \log a \cdot \left(1 - \frac{1}{t^2}ight)$.Next,differentiate $x$ with respect to $t$ using the power rule:$\frac{dx}{dt} = \frac{d}{dt}\left(t+\frac{1}{t}ight)^{a} = a\left(t+\frac{1}{t}ight)^{a-1} \cdot \frac{d}{dt}\left(t+\frac{1}{t}ight) = a\left(t+\frac{1}{t}ight)^{a-1} \cdot \left(1 - \frac{1}{t^2}ight)$.Now,find $\frac{dy}{dx}$ using the parametric form formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:$\frac{dy}{dx} = \frac{a^{t+\frac{1}{t}} \cdot \log a \cdot \left(1 - \frac{1}{t^2}ight)}{a\left(t+\frac{1}{t}ight)^{a-1} \cdot \left(1 - \frac{1}{t^2}ight)}$.Canceling the common term $\left(1 - \frac{1}{t^2}ight)$ (for $t 
eq \pm 1$):$\frac{dy}{dx} = \frac{a^{t+\frac{1}{t}} \log a}{a\left(t+\frac{1}{t}ight)^{a-1}}$.

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

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