Question

Please solve the question

$y = {e}^{ {x}^{ {e}^{ {x}^{ {e}^{x - - - \infty } } } } } \: find \: \frac{dy}{dx}$
​

Answer 1

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

Given function is

$\rm :\longmapsto\:y = {e}^{ {x}^{ {{e}^{x}}^{{}^{} - - - \infty } } }$

can be rewritten as

$\rm :\longmapsto\:y = {e}^{ {x}^{y} } - - - (1)$

On taking log on both sides, we get

$\rm :\longmapsto\:logy = log{e}^{ {x}^{y} }$

$\rm :\longmapsto\:logy = {x}^{y}loge$

$\rm :\longmapsto\:logy = {x}^{y} - - - (2)$

On taking log on both sides, we get

$\rm :\longmapsto\:log(logy) = log {x}^{y}$

$\rm :\longmapsto\:log(logy) = ylogx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}log(logy) = \dfrac{d}{dx}ylogx$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}logx = \frac{1}{x} \: }}}$

and

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}uv = \:u\dfrac{d}{dx}v + v\dfrac{d}{dx}u \: }}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{d}{dx}logy = y\dfrac{d}{dx}logx + logx\dfrac{d}{dx}y$

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{1}{logy}\dfrac{dy}{dx} = y \times \dfrac{1}{x} + logx\dfrac{dy}{dx}$

$\rm :\longmapsto\:\dfrac{1}{ylogy}\dfrac{dy}{dx} = \dfrac{y}{x} + logx\dfrac{dy}{dx}$

$\rm :\longmapsto\:\dfrac{1}{ylogy}\dfrac{dy}{dx} - logx\dfrac{dy}{dx} = \dfrac{y}{x}$

$\rm :\longmapsto\:\bigg[\dfrac{1}{ylogy} - logx\bigg]\dfrac{dy}{dx} = \dfrac{y}{x}$

$\rm :\longmapsto\:\bigg[\dfrac{1 - y \: logy \: logx}{ylogy}\bigg]\dfrac{dy}{dx} = \dfrac{y}{x}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ {y}^{2} \: logy }{x(1 - y \: logy \: logx)}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

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

Given function is

$\rm :\longmapsto\:y = {e}^{ {x}^{ {{e}^{x}}^{{}^{} - - - \infty } } }$

can be rewritten as

$\rm :\longmapsto\:y = {e}^{ {x}^{y} } - - - (1)$

On taking log on both sides, we get

$\rm :\longmapsto\:logy = log{e}^{ {x}^{y} }$

$\rm :\longmapsto\:logy = {x}^{y}loge$

$\rm :\longmapsto\:logy = {x}^{y} - - - (2)$

On taking log on both sides, we get

$\rm :\longmapsto\:log(logy) = log {x}^{y}$

$\rm :\longmapsto\:log(logy) = ylogx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}log(logy) = \dfrac{d}{dx}ylogx$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}logx = \frac{1}{x} \: }}}$

and

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}uv = \:u\dfrac{d}{dx}v + v\dfrac{d}{dx}u \: }}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{d}{dx}logy = y\dfrac{d}{dx}logx + logx\dfrac{d}{dx}y$

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{1}{logy}\dfrac{dy}{dx} = y \times \dfrac{1}{x} + logx\dfrac{dy}{dx}$

$\rm :\longmapsto\:\dfrac{1}{ylogy}\dfrac{dy}{dx} = \dfrac{y}{x} + logx\dfrac{dy}{dx}$

$\rm :\longmapsto\:\dfrac{1}{ylogy}\dfrac{dy}{dx} - logx\dfrac{dy}{dx} = \dfrac{y}{x}$

$\rm :\longmapsto\:\bigg[\dfrac{1}{ylogy} - logx\bigg]\dfrac{dy}{dx} = \dfrac{y}{x}$

$\rm :\longmapsto\:\bigg[\dfrac{1 - y \: logy \: logx}{ylogy}\bigg]\dfrac{dy}{dx} = \dfrac{y}{x}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ {y}^{2} \: logy }{x(1 - y \: logy \: logx)}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$