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If ey(x+1)=1e^y(x+1)=1, then

Solution

✅ Correct Option: 1

From ey(x+1)=1e^y(x+1)=1 we get y=−log⁡e(x+1)y = -\log_e(x+1).

Then dydx=−1x+1\dfrac{dy}{dx} = \dfrac{-1}{x+1} and d2ydx2=1(x+1)2\dfrac{d^2y}{dx^2} = \dfrac{1}{(x+1)^2}.

Since (dydx)2=1(x+1)2\left(\dfrac{dy}{dx}\right)^2 = \dfrac{1}{(x+1)^2}, we have d2ydx2=(dydx)2\dfrac{d^2y}{dx^2} = \left(\dfrac{dy}{dx}\right)^2.

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