If e^y\left ( x+1 \right )=1, then show that \frac{\mathrm{d^2}y }{\mathrm{d} x^2}=\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}

 

 

 

 
 
 
 
 

Answers (1)

e^y\left ( x+1 \right )=1,

By differentiating both sides wrt x

e^y\times 1+(x+1)\times \frac{e^ydy}{dx}=0

 

\frac{e^ydy}{dx}\times (x+1)=-e^{y}

\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{-e^y}{e^y.(x+1)}\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}= \frac{-1}{x+1}..........(1)

Differentiating again wrt x

\frac{\mathrm{d}^2 y}{\mathrm{d} x^2}=-(-1)(x+1)^{-2}.1\Rightarrow \frac{1}{(x+1)^2}\Rightarrow \left \{ \frac{-1}{x+1} \right \}^2 ...............(2)

 

\frac{\mathrm{d^2}y }{\mathrm{d} x^2}=\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}

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