A) \[\frac{x}{1+x}\]
B) \[\frac{1}{x}\]
C) \[\frac{1-x}{x}\]
D) \[\frac{1+x}{x}\]
Correct Answer: C
Solution :
\[x={{e}^{y+{{e}^{y+.....\infty }}}}\] \[\therefore \]\[x={{e}^{y+x}}\] Taking log on both sides, we get \[\log x=(y+x)\] Differentiate w.r. to\[x,\]we get \[\frac{1}{x}=\frac{dy}{dx}+1\Rightarrow \frac{dy}{dx}=\frac{1-x}{x}\]You need to login to perform this action.
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