Question

If y=a^x^a^x^a^x... , then prove that dy/dx= (y^2*logy)/x(1-y*logx logy)​

Answer 1

Topic :-

Differentiation

To Prove :-

$If\:y=a^{x^{a^{x^{a^{\cdot^\cdot}}}}}\:where\:'a'\:is\:constant,then$

$\dfrac{dy}{dx}=\dfrac{y^2\ln y}{x(1-y\ln x\ln y)}$

Solution :-

Generally, in questions where we have to find derivative of a function which involves variable power, we take logarithm of the function and then differentiate.

$y=a^{x^{a^{x^{a^{\cdot^\cdot}}}}}$

Taking 'log' both sides,

$\ln y=\ln a^{x^{a^{x^{a^{\cdot^\cdot}}}}}$

$\ln y={x^{a^{x^{a^{\cdot^\cdot}}}}}\ln a$

$(\because \ln p^q=q\ln p)$

Taking 'log' both sides again,

$\ln(\ln y)=\ln\left({x^{a^{x^{a^{\cdot^\cdot}}}}}\ln a\right)$

$\ln(\ln y)=\ln\left({x^{a^{x^{a^{\cdot^\cdot}}}}}\right)+\ln(\ln a)$

$(\because\ln pq=\ln p+\ln q)$

$\ln(\ln y)=\left({a^{x^{a^{\cdot^\cdot}}}}\right)\ln x+\ln(\ln a)$

$(\because \ln p^q=q\ln p)$

$\ln(\ln y)=y\ln x+\ln(\ln a)$

$\left(\because y=a^{x^{a^{x^{a^{\cdot^\cdot}}}}} \right)$

Differentiating,

$\dfrac{d}{dx}\left(\ln(\ln y)\right)=\dfrac{d}{dx}\left(y\ln x\right)+\dfrac{d}{dx}\left(\ln(\ln a)\right)$

$\left(\dfrac{1}{\ln y}\right)\cdot\left(\dfrac{d}{dx}\left(\ln y\right)\right)=\dfrac{d}{dx}\left(y\ln x\right)+\dfrac{d}{dx}\left(\ln(\ln a)\right)$

$\left(\because \dfrac{d}{dx}\left(\ln t\right)=\dfrac{1}{t}\cdot\dfrac{dt}{dx} \right)$

$\left(\dfrac{1}{\ln y}\right)\cdot\left(\dfrac{1}{y}\right)\cdot\left(\dfrac{dy}{dx}\right)=\dfrac{d}{dx}\left(y\ln x\right)+\dfrac{d}{dx}\left(\ln(\ln a)\right)$

$\left(\because \dfrac{d}{dx}\left(\ln t\right)=\dfrac{1}{t}\cdot\dfrac{dt}{dx} \right)$

$\left(\dfrac{1}{y\ln y}\right)\cdot\left(\dfrac{dy}{dx}\right)=\left(\ln x\cdot\dfrac{dy}{dx}+y\cdot\dfrac{d}{dx}\left(\ln x \right)\right)+\dfrac{d}{dx}\left(\ln(\ln a)\right)$

$\left(\because \dfrac{d(fg)}{dx}=g\cdot\dfrac{df}{dx}+f\cdot\dfrac{dg}{dx}\right)$

$\left(\dfrac{1}{y\ln y}\right)\cdot\left(\dfrac{dy}{dx}\right)=\left(\ln x\cdot\dfrac{dy}{dx}+y\cdot\dfrac{1}{x}\right)+\dfrac{d}{dx}\left(\ln(\ln a)\right)$

$\left(\because \dfrac{d}{dx}\left(\ln x\right)=\dfrac{1}{x} \right)$

$\left(\dfrac{1}{y\ln y}\right)\cdot\left(\dfrac{dy}{dx}\right)=\left(\ln x\cdot\dfrac{dy}{dx}+\dfrac{y}{x}\right)+0$

$\left(\because \dfrac{dk}{dx}=0,when\:k\:is\:constant. \right)$

Grouping terms,

$\left(\dfrac{1}{y\ln y}\right)\cdot\left(\dfrac{dy}{dx}\right)-\ln x\cdot\dfrac{dy}{dx}=\dfrac{y}{x}$

$\left(\dfrac{dy}{dx}\right)\cdot\left(\dfrac{1}{y\ln y}-\ln x\right)=\dfrac{y}{x}$

Taking LCM and solving,

$\left(\dfrac{dy}{dx}\right)\cdot\left(\dfrac{1-y\ln x\ln y}{y\ln y}\right)=\dfrac{y}{x}$

$\dfrac{dy}{dx}=\dfrac{y}{x}\left(\dfrac{y\ln y}{1-y\ln x\ln y}\right)$

$\underline{\boxed{\dfrac{dy}{dx}=\dfrac{y^2\ln y}{x(1-y\ln x\ln y)}}}$

Hence, Proved !!