ln(x + y) = 4xy

For x = 0

lny = 0

⟹ y = 1

ln(x + y) = 4xy

⟹ \(\frac{1}{x+y}(1 + \frac{dy}{dx})\)

= 4y + 4x\(\frac{dy}{dx}\)  .....(1)

At x = 0 and y = 1

⟹ \(\frac{1}{1}\)\(\Big(1 +\frac{dy}{dx}\Big)\) = 4 + 0

⟹ \(\frac{dy}{dx}|_{x=0}\) = 3

Differentiate Equation (1)

= \(\frac{1}{x+y}(1 + \frac{dy}{dx})\)

= 4y + 4x\(\frac{dy}{dx}\)

⟹ \(\frac{-1}{(x+y)^2}\Big(1+ \frac{dy}{dx}\Big)^2\) + \(\Big(\frac{1}{x+y}\Big)\)\(\Big(\frac{d^2y}{dx^2}\Big)\)

= \(4\frac{dy}{dx}\) + \(4\frac{dy}{dx}\) + \(4x\frac{d^2y}{dx^2}\)

Put x = 0, y = 1 and \(\frac{dy}{dx}\) = 3

⟹ \(\frac{-1}{(0+1)^2}(1+3)^2 + \Big( \frac{1}{0+1}\Big)\)\(\Big(\frac{d^2y}{dx^2}\Big)\)

= 4x³ + 4 x 3 + 0

⟹ \(-16 +\frac{d^2y}{dx^2}\) = 24

⟹ \(\frac{d^2y}{dx^2}\) = 40