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

JamiaJamia 2011

Solution:

Given, $ x={{e}^{y+x}} $ $ \Rightarrow $ $ log\text{ }x=x+y $ On differentiating w.r.t. $ x, $ we get $ \frac{1}{x}=1+\frac{dy}{dx} $ $ \Rightarrow $ $ \frac{dy}{dx}=\frac{1-x}{x} $