If x = e^ y + e^ y + ....t o , , , x > 0, then dy dx is - Vidyadip

MCQ

If $x = {e^{y + {e^{y + ....t{\rm{o}}\,\,\infty }}}}$, $x > 0,$ then ${{dy} \over {dx}}$ is

A
${{1 + x} \over x}$
B
${1 \over x}$
✓
${{1 - x} \over x}$
D
${x \over {1 + x}}$

✓

Answer

Correct option: C.

${{1 - x} \over x}$

c
(c) $x = {e^{y + {e^{y + ....to\,\infty }}}}$, $x > 0$ $\Rightarrow$ $x = {e^{y + x}}$

Taking log to the both sides, $\log x = (y + x)$

Differentiate both sides w.r.t. $x,$ $\frac{1}{x} = \frac{{dy}}{{dx}} + 1$

$ \Rightarrow \frac{{dy}}{{dx}} = \frac{{1 - x}}{x}$.

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