Question

$\huge \mathcal \red{Question}$​

Answer 1

Answer:

$\dag \sf \red{x^y = e^x-^y}$

Question no. 1

• Find dy/dx at x = 1

Answer :-

taking log both sides

$logx^y=loge^x-y$

$ylogx=x-yloge$

differentiate both sides w.r.t x

1/y dy/DX=1-dy/DX (loge=1)

1/y dy/DX +dy/DX =1

dy/DX(1/y+1)=1

dy/DX(1+y/y)=1

dy/DX=

y/1+y answer

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Question no. 2

• Find d²y/d²x at x = 1

Answer :-

Given:

          $x^y = e^{x-y}$

To Find:

         $\frac{dy}{dx} = \frac{ log x}{(1 + log x)^2}$

Solution:

      ⇒           $x^y = e^{x-y}$

taking log on both sides

      ⇒       $log(x^y )= log( e^{x-y})$

      ⇒      $(y) log \ x = (x-y) log(e )$

      ⇒       $y log\ x = x - y$                                  $[ log \ e =1 ]$

      ⇒     $y \ log\ x + y = x$

      ⇒     $y ( log x + 1 ) = x$

      ⇒      $y = \frac{x}{1 + log \ x}$

On differentiating both sides with respect to x

 ⇒          

$\frac{dy}{dx} = \frac{(1+ log \ x) \frac{d}{dx} (x)- x \frac{d}{dx} (1+log \ x)}{(1+ log \ x)^2}$

 ⇒          $\frac{dy}{dx} = \frac{1 + log \ x - (x) \frac{1}{x} }{(1 + log \ x)^2}$

 ⇒          $\frac{dy}{dx} = \frac{1 + log \ x - 1 }{(1 + log \ x)^2}$

 ⇒          $\frac{dy}{dx} = \frac{ log x}{(1 + log x)^2}$

Hence proved✓

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Answer 2

Step-by-step explanation:

Solution:

⇒ (y) log \ x = (x-y) log(e )(y)log x=(x−y)log(e)