JEE MAIN - Mathematics (2004 - No. 31)
If $$x = {e^{y + {e^y} + {e^{y + .....\infty }}}}$$ , $$x > 0,$$ then $${{{dy} \over {dx}}}$$ is
$${{1 + x} \over x}$$
$${1 \over x}$$
$${{1 - x} \over x}$$
$${x \over {1 + x}}$$
Explanation
$$x = {e^{y + {e^{y + .....\infty }}}}\,\, \Rightarrow x = {e^{y + x}}.$$
Taking log.
$$\log \,\,x = y + x$$
$$ \Rightarrow {1 \over x} = {{dy} \over {dx}} + 1$$
$$ \Rightarrow {{dy} \over {dx}} = {1 \over x} - 1 = {{1 - x} \over x}$$
Taking log.
$$\log \,\,x = y + x$$
$$ \Rightarrow {1 \over x} = {{dy} \over {dx}} + 1$$
$$ \Rightarrow {{dy} \over {dx}} = {1 \over x} - 1 = {{1 - x} \over x}$$
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