Question

If \(e^y(x+1)=1\),show that \(\frac{d^2y}{dx^2}=(\frac{dy}{dx})^2\)

Answer 1

The given relationship is,\(e^y(x+1)=1\)
\(e^y(x+1)=1\)
\(⇒e^y=\frac{1}{x+1}\)
Taking logarithm on both the sides, we obtain
\(y=log\frac{1}{(x+1)}\)
Differentiating this relationship with respect to \(x\), we obtain
\(\frac{dy}{dx}=(x+1)\frac{d}{dx}(\frac{1}{x+1})=(x+1).\frac{-1}{(x+1)^2}=\frac{-1}{(x+1)}\)
\(∴\frac{d^2y}{dx^2}=\frac{-d}{dx}(\frac{1}{(x+1)})=-(\frac{-1}{(x+1)^2})\)\(=\frac{1}{(x+1)^2}\)
\(⇒\frac{d^2y}{dx^2}=(\frac{-1}{(x+1)})^2\)
\(\frac{d^2y}{dx^2}=(\frac{dy}{dx})^2\)
Hence, proved.