Aspire Faculty ID #18333 · Topic: AMU MCA 2017 · Just now
AMU MCA 2017

If $y=e^{x}+e^{x+e^{x}}+e^{x+e^{x+e^{x}}}+\dots$, then $\frac{dy}{dx}$ is equal to

Solution

$y=e^{x+y}$ Taking log, $\log y=x+y$ Differentiating, $\frac{1}{y}\frac{dy}{dx}=1+\frac{dy}{dx}$ $\frac{dy}{dx}\left(\frac{1}{y}-1\right)=1$ $\frac{dy}{dx}=\frac{y}{1-y}$

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