Practicing Success
Practicing Success
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If $x=e^{y+e^{y+e^{y+... ~to ~\infty}}}, x > 0$, then $\frac{dy}{dx}$, is |
$\frac{1+x}{x}$ $\frac{1}{x}$ $\frac{1-x}{x}$ $\frac{x}{1+x}$ |
$\frac{1-x}{x}$ |
We have, $x=e^{y+e^{y+e^{y+... ~to~\infty}}}$ $\Rightarrow x=e^{y+x}$ $\Rightarrow \log _e x=(y+x)$ $\Rightarrow \frac{1}{x}=\frac{d y}{d x}+1$ [Differentiating with respect to x] $\Rightarrow \frac{d y}{d x}=\frac{1-x}{x}$ |
