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Find The Derivative of
$\huge\pmb{x {}^{1/x} }$


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Answer 1

Answer:

Let $y = x^{\frac{1}{x} }$

Take log on both side

$log(y) = \frac{1}{x}log(x)$

Now differentiate bothe side w.r.t x

$\frac{1}{y} \frac{dy}{dx}= \frac{1-logx}{x^{2} }$

$\frac{dy}{dx} = y(\frac{1-logx}{x^{2} } )$

$\frac{dy}{dx} = x^{\frac{1}{x} } (\frac{1-logx}{x^{2} } )$

This is your ans

Answer 2

• Let The Given Function be = y

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▪ Given :-

$\large \bf{y =  {x}^{1/x }}$

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▪ To Find :-

$\purple{  \bf\large \dfrac{dy}{dx} }$

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▪ Solution :-

We Have ,

$\large \mathtt{y = x {}^{1/x} }$

$\large \bigstar \underline{ \pmb{ \mathfrak{  \text{T}a \text king    \:  \: \text Log \:    \: Both  \:  \:  \text Side  }}}$

$:  \longmapsto   \sf \log y  =  \log  { \mathtt x}^{1/ \mathtt x}  \\  \\ :  \longmapsto \sf \log y =  \frac{1}{ \mathtt x}  \log \mathtt x \\  \bf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \{ \because  log {m}^{n}  = n. log m \}$

$\large\bigstar \underline{ \pmb{ \mathfrak{ Differentiating\:both\: sides\: \text{w.r.t x}  }}}$

$\small:\longmapsto  \sf\frac{1}{y} . \frac{dy}{d \mathtt x}  =  \frac{1}{ \mathtt x} . \frac{d}{d\mathtt x} ( \log  \mathtt x)+  \log \mathtt x. \frac{d}{d\mathtt x} (x {}^{ -1} ) \\  \bf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \{ \because  Product\:\:Rule \}\\  \\ :\longmapsto \sf \frac{1}{y}  . \frac{dy}{d\mathtt x}  =  \frac{1}{\mathtt x} . \frac{1}{\mathtt x}  +  \log \mathtt x. \bigg( -  \frac{  1}{ {\mathtt x}^{2} }  \bigg ) \\  \\ : \longmapsto \sf \frac{1}{y} . \frac{dy}{d\mathtt x}  =  \frac{1}{ {\mathtt x}^{2}  }  -  \frac{ \log\mathtt x}{ {\mathtt x}^{2} }  \\  \\ : \longmapsto \sf \frac{dy}{d\mathtt x}  = y \bigg( \frac{1 -  \log\mathtt x}{ {\mathtt x}^{2} }  \bigg) \\  \\ \large\purple{ : \longmapsto \pmb{ \underline {\boxed{{ \frac{dy}{dx}  =  \frac{x {}^{1/x} }{ {x}^{2}  } \bigg(  1 -  \log x\bigg)} }}}}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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