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.