Question

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


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Answer 1

$\underline{\huge{Answer:-}}$

Step-by-step explanation:

$\underline{\huge{Solution:-}}$

Observe that both y and x are defined for all real t $\: \neq \:$ 0. Clearly

$\implies \: \large\mathcal{ \frac{dy}{dt} = \frac{d}{dt}( {a}^{t + \frac{1}{t} }) = {a}^{t + \frac{1}{t} } \frac{d}{dt}(t + \frac{1}{t}). \: log \: a } \\ \\ \implies \large \mathcal{ = {a}^{t + \frac{1}{t} } (1 - \frac{1}{ {t}^{2} })log \: a }$

Sɪᴍɪʟᴀʀʟʏ,

$\implies \large \mathcal{ \frac{dx}{dt} = a[t + \frac{1}{t} ] ^{a - 1} . \frac{d}{dt}(t + \frac{1}{t}) } \\ \\ \implies \large \mathcal{a[t + \frac{1}{t} ] ^{a - 1}.(1 - \frac{1}{ {t}^{2} }) }$

$\large \mathcal{ \frac{dx}{dt} \neq0 \: only \: if \: \neq \: \pm \: 1. \: Thus \: for \: t \: \neq \: \pm \: 1, }$

$\implies \large \mathcal{ \frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } = \frac{ {a}^{t + \frac{1}{t} }(1 - \frac{1}{ {t}^{2} })log \: a }{a [t + \frac{1}{t} ] ^{a - 1} .(1 - \frac{1}{ {t}^{2} } )} } \\ \\ \: \: \: \: \Large \fbox \mathcal{Ans = \frac{ {a}^{t + \frac{1}{t} }log \: a }{a(t + \frac{1}{t} ) ^{a - 1} } }$