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If $y=e^{{x + e}^{{x + e}^{{x+....+∞}}}}$ then $\frac{dy}{dx}=$ |
$\frac{x}{1-x}$ $\frac{1-x}{x}$ $\frac{y}{1-y}$ $\frac{1-y}{y}$ |
$\frac{y}{1-y}$ |
The correct answer is Option (3) → $\frac{y}{1-y}$ $y=e^{{x + e}^{{x + e}^{{x+e}^{...+∞}}}}=y=e^{x+y}$ $\log y = x+y$ differentiating wrt x $\frac{1}{y}\frac{dy}{dx}=1+\frac{dy}{dx}$ so $\frac{dy}{dx}=\frac{y}{1-y}$ |
