Math, asked by parineetasakshi6475, 1 year ago

If x^y = e^x – y, prove that dy/dx = log x/(1 + log x)^2

Answers

Answered by roshinik1219
19

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

Answered by sandy1816
8

Answer:

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