Question

Find dy/dx
when:
dx
y = x power 1/x
​

Answer 1

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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 ,

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

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

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

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

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

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

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Answer 2

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

$\large \underline{ \pmb{ \mathcal{ \text{T}a \text king \: \: \text Log \: \: Both \: \: \text Side \: \: }}} : -$

$: \implies \sf \log y = \log { \mathtt x}^{1/ \mathtt x} \\ \\ : \implies \sf \log y = \frac{1}{ \mathtt x} \log \mathtt x$

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

$: \implies \sf\dfrac{1}{y} . \dfrac{dy}{d \mathtt x} = \frac{1}{ \mathtt x} . \frac{d}{d\mathtt x} ( \log \mathtt x)+ \log \mathtt x. \frac{d}{d\mathtt x} (x {}^{ -1} ) \\ \\ : \implies \sf \frac{1}{y} . \frac{dy}{d\mathtt x} = \frac{1}{\mathtt x} . \frac{1}{\mathtt x} + \log \mathtt x. \bigg( - \frac{ 1}{ {\mathtt x}^{2} } \bigg ) \\ \\ : \implies \sf \frac{1}{y} . \frac{dy}{d\mathtt x} = \frac{1}{ {\mathtt x}^{2} } - \frac{ \log\mathtt x}{ {\mathtt x}^{2} } \\ \\ : \implies \sf \frac{dy}{d\mathtt x} = y \bigg( \frac{1 - \log\mathtt x}{ {\mathtt x}^{2} } \bigg) \\ \\ \red{ : \implies \bf{ \underline {\boxed{{ \bf\frac{dy}{dx} = \frac{x {}^{1/x} }{ {x}^{2} } \bigg( 1 - \log x\bigg)} }}}}$