Question

find dy/dx if x=(t+1/t)^a, y=a^t+1/t, a> 0,
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\red{\rm :\longmapsto\:x = {\bigg(t + \dfrac{1}{t} \bigg) }^{a}}$

and

$\red{\rm :\longmapsto\:y = {\bigg(a \bigg) }^{\bigg(t + \dfrac{1}{t} \bigg)}}$

Now, Consider

$\red{\rm :\longmapsto\:x = {\bigg(t + \dfrac{1}{t} \bigg) }^{a}}$

On differentiating both sides w. r. t. t, we get

${\rm :\longmapsto\:\dfrac{d}{dt}x = \dfrac{d}{dt} {\bigg(t + \dfrac{1}{t} \bigg) }^{a}}$

$\red{\boxed{ \rm{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}}$

${\rm :\longmapsto\:\dfrac{dx}{dt} = a {\bigg(t + \dfrac{1}{t} \bigg) }^{a - 1}}\dfrac{d}{dt}\bigg(t + \dfrac{1}{t} \bigg)$

$\rm :\longmapsto\:\dfrac{dx}{dt} = a {\bigg(t + \dfrac{1}{t} \bigg) }^{a - 1}\bigg(1 - \dfrac{1}{ {t}^{2} } \bigg)$

Now, Consider

$\red{\rm :\longmapsto\:y = {\bigg(a \bigg) }^{\bigg(t + \dfrac{1}{t} \bigg)}}$

On differentiating w. r. t. t, we get

${\rm :\longmapsto\:\dfrac{d}{dt}y = \dfrac{d}{dt} {\bigg(a \bigg) }^{\bigg(t + \dfrac{1}{t} \bigg)}}$

$\red{\boxed{ \rm{ \dfrac{d}{dx} {a}^{x} = {a}^{x}loga}}}$

${\rm :\longmapsto\:\dfrac{dy}{dt} = {\bigg(a \bigg) }^{\bigg(t + \dfrac{1}{t} \bigg)}loga \: \dfrac{d}{dt}\bigg(t + \dfrac{1}{t} \bigg)}$

$\rm :\longmapsto\:\dfrac{dy}{dt} = {\bigg(a \bigg) }^{\bigg(t + \dfrac{1}{t} \bigg)}loga \: \bigg(1 - \dfrac{1}{ {t}^{2} } \bigg)$

Now, Consider

$\red{\rm :\longmapsto\:\dfrac{dy}{dx}}$

$\rm \: = \: \: \dfrac{dy}{dt} \: \div \: \dfrac{dx}{dt}$

$\rm \: = \: \: \dfrac{{\bigg(a \bigg) }^{\bigg(t + \dfrac{1}{t} \bigg)}loga \: \bigg(1 - \dfrac{1}{ {t}^{2} } \bigg)}{a {\bigg(t + \dfrac{1}{t} \bigg) }^{a - 1}\bigg(1 - \dfrac{1}{ {t}^{2} } \bigg)}$

$\rm \: = \: \: \dfrac{{\bigg(a \bigg) }^{\bigg(t + \dfrac{1}{t} \bigg)}loga \: }{a {\bigg(t + \dfrac{1}{t} \bigg) }^{a - 1} }$

Additional Information :-

$\boxed{ \rm{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}$

$\boxed{ \rm{ \dfrac{d}{dx}k = 0}}$

$\boxed{ \rm{ \dfrac{d}{dx}x = 1}}$

$\boxed{ \rm{ \dfrac{d}{dx}sinx = cosx}}$

$\boxed{ \rm{ \dfrac{d}{dx}cosx = - \: sinx}}$

$\boxed{ \rm{ \dfrac{d}{dx}cosecx = - \: cosecx \: cotx}}$

$\boxed{ \rm{ \dfrac{d}{dx}secx = \: secx \: tanx}}$

$\boxed{ \rm{ \dfrac{d}{dx}tanx = \: sec ^{2} x \:}}$

$\boxed{ \rm{ \dfrac{d}{dx}cotx = - \: cosec ^{2} x \:}}$