Question

Differentiate the following function with respect to x

${e}^{ {x}^{ {e}^{x} } }$
​

Answer 1

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

Given function is

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

Taking log on both sides, we get

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

We know,

$\red{\rm :\longmapsto\:\boxed{\tt{ log {x}^{y} = y \: logx \: }}}$

So, using this, we get

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

$\rm :\longmapsto\:logy = {x}^{ {e}^{x} } \: \: \: \{ \because \: loge = 1 \} - - - (2)$

Taking log on both sides, we get

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

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

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

$\rm :\longmapsto\:\dfrac{d}{dx} log(logy) = \dfrac{d}{dx} \bigg[ {e}^{x} \: logx \: \bigg]$

We know,

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

and

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

So, using this, we get

$\rm :\longmapsto\:\dfrac{1}{logy}\dfrac{d}{dx} logy = logx\dfrac{d}{dx} {e}^{x} + {e}^{x}\dfrac{d}{dx} logx$

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

$\rm :\longmapsto\:\dfrac{dy}{dx} = y \: logy \: \bigg[{e}^{x}logx + \dfrac{{e}^{x}}{x}\bigg]$

On substituting the values of y and logy from equation (1) and (2), we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {x}^{ {e}^{x} }\:{e}^{ {x}^{ {e}^{x} } } \: \bigg[{e}^{x}logx + \dfrac{{e}^{x}}{x}\bigg]$

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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

Step-by-step explanation:

$\mathtt \purple{{e}^{ {x}^{ {e}^{x} } }}$