If [x = e^y + e^y + e^y + ..... ] , x 0, then [dy dx] is

Question

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

Accepted Answer

$$x = {e^{y + {e^{y + .....\infty }}}}\,\, \Rightarrow x = {e^{y + x}}.$$

Taking log.

$$\log \,\,x = y + x$$

$$ \Rightarrow {1 \over x} = {{dy} \over {dx}} + 1$$

$$ \Rightarrow {{dy} \over {dx}} = {1 \over x} - 1 = {{1 - x} \over x}$$

Solutions · 1