If x=e^{y+e^{y+..to\; \infty }},x> 0\; then\; \frac{dy}{dx}\; \;    is

  • Option 1)

    \frac{1-x}{x}\;

  • Option 2)

    \; \frac{1}{x}\;

  • Option 3)

    \; \frac{x}{1+x}\;

  • Option 4)

    \; \frac{1+x}{x}

 

Answers (1)

As we learnt in 

Differential Equations -

An equation involving independent variable (x), dependent variable (y) and derivative of dependent variable with respect to independent variable 
\left (\frac{\mathrm{d} y}{\mathrm{d} x} \right )

- wherein

eg:

  \frac{d^{2}y}{dx^{2}}- 3\frac{dy}{dx}+5x=0

 

 x=e^{y+}^{e^{y}.......\infty}

x=e^{y+x}

logx=(x+y)

\frac{1}{x}=1+\frac{dy}{dx}

\frac{dy}{dx}=\frac{1-x}{x}

 


Option 1)

\frac{1-x}{x}\;

This option is correct

Option 2)

\; \frac{1}{x}\;

This option is incorrect

Option 3)

\; \frac{x}{1+x}\;

This option is incorrect

Option 4)

\; \frac{1+x}{x}

This option is incorrect

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