Question

If e^y(x+1)=1, show that d^2/dx^2=(dy/dx)^2

Answer 1

Given, $\bf{e^y(x+1)=1}$
we have to proof :$\bf{\frac{d^2y}{dx^2}=\left(\begin{array}{c}\frac{dy}{dx}\end{array}\right)^2}$

e^y (x + 1) = 1
=> e^y = 1/(x + 1)
take log both sides,
=> y = log{1/(x + 1)}
differentiate y with respect to x,
=> dy/dx = d[log{1/(x + 1)}]/dx
=> dy/dx = 1/{1/(x + 1)} × -1/(x +1)²
=> dy/dx = (x + 1)/1 × -1/(x +1)²
=> dy/dx = -1/(x +1) ------(1)
squaring both sides,
(dy/dx)² = 1/(x + 1)² --------(2)

from equation (1),
dy/dx = -1/(x +1) = $\bf{-(x+1)^{-1}}$
differentiate once again,
d²y/dx² = $\bf{-1\times(-1)(x+1)^{-1-1}}$
=> d²y/dx² = $\bf{(x + 1)^{-2}$
e.g., d²y/dx² = 1/(x + 1)² -----(3)

from equations (2) and (3),
d²y/dx² = (dy/dx)² $\textbf{\underline{hence proved}}$