Practicing Success
Practicing Success
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If $x=e^{y+e^{y+e^{y+... \infty}}}$, x > 0, then $\frac{d y}{d x}$ is equal to |
$\frac{x}{1+x}$ $\frac{1}{x}$ $\frac{1-x}{x}$ $\frac{1+x}{x}$ |
$\frac{1-x}{x}$ |
$x=e^{y+e^{y+e^{y+... \infty}}}$ ⇒ $\log x = y + e^{y+e^{y+e^{y+... \infty}}}$ ⇒ $\log x = y + x$ differentiating wrt x $\frac{1}{x} = \frac{dy}{dx} + 1 ⇒ \frac{dy}{dx} = \frac{1-x}{x}$ Option: 3 |
