If x=ey+ey+ey........∞ and dydx=1−kxx, then k=? - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

Given $x = e^{y + x}$

Take logarithm on both sides

$\begin{array}{l} \log x = \log e^{y + x} \\ \log x = (y + x) \log e \\ \log x = y + x \end{array}$

Differentiate both sides with respect to $x$

$\frac{1}{x} = \frac{d y}{d x} + 1 \Rightarrow \frac{d y}{d x} = \frac{1}{x} - 1 \Rightarrow \frac{d y}{d x} = \frac{1 - x}{x}$

Therefore, $k = 1$

If $x = e^{y + e^{y + e^{y ........ \infty}}}$ and $\frac{d y}{d x} = \frac{1 - k x}{x}, then k = ?$