Hindi, asked by kunjtrambadia9, 19 hours ago

चित्र के आधार पर चारित्रिक विशेषताओं व _______ का वर्णन करना चाहिए​

Answers

Answered by vimaljegi
0

Explanation:

Let The Given Function be = y

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▪ Given :-

\large \bf{y =  {x}^{1/x }}y= x1/x

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▪ To Find :-

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

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▪ Solution :-

We Have ,

\large \mathtt{y = x {}^{1/x} }y=x1/x

\begin{gathered} \large \bigstar \underline{ \pmb{ \mathfrak{  \text{T}a \text king    \:  \: \text Log \:    \: Both  \:  \:  \text Side  }}} \\ \end{gathered}★ Taking    Log   Both   Side  Taking    Log   Both   Side 

\begin{gathered} :  \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 \}\end{gathered}: ⟼  logy = log x1/x  : ⟼logy= x1 logx             {∵ logmn =n.logm}

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

\begin{gathered}\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)} }}}}\end{gathered}:⟼ y1.dxdy = x1.dxd(log x)+ logx.dxd(x−1)             {∵ ProductRule} :⟼y1 .dxdy = x1.x1 + logx.(− x2 1 ) :⟼y1.dxdy = x2 1 − x2logx  :⟼dxdy =y(x21− logx ) :⟼dxdy = x2 x1/x( 1− logx)dxdy = x2 x1/x( 1− logx)

\begin{gathered} \Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}\end{gathered} Which  is  the  required

Answered by ritexa93
1

Answer:

Tara question no jawab.

lakh phata phat

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