Explanation

(b)
Given, $x = log_{e}t$
$\Rightarrow \ e^{x} = \ t\ and\ \ y + 1 = t^{2} \Rightarrow \ y = e^{2x} - 1\ \$
On differentiating w.r.t. $y$, we get
$2e^{2x}\frac{dx}{dy} = 1$
$\Rightarrow \ \ \frac{dx}{dy} = \frac{1}{2e^{2x}}$
Again, differentiating w.r.t. $y,\$we get
$\frac{d^{2}x}{dy^{2}} = \frac{1}{2}e^{- 2x}( - 2)\frac{dx}{dy}$
$= - e^{- 2x}.\frac{1}{2e^{2x}}$
$= - \frac{1}{2}\ e^{- 4x}$