Question

If $y=y\left(x\right)$ is an implicit function of $x$ such that ${\log }_e(x+y)=4xy,$ then $\frac{d^{2}y}{d{x}^2}$ at $x=0$ is equal to

Answer 1

${\log }_e(x+y)=4xy$ at x = 0, y = 1
$\frac{1}{x+y}[1+\frac{dy}{dx}]=4[y+x\cdot \frac{dy}{dx}]$
$\Rightarrow \frac{1}{x+y}-4y=(4x-\frac{1}{x+y})\frac{dy}{dx}$
$\Rightarrow 1-4xy-4y^2=(4x^2+4xy-1)\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{1-4xy-4y^2}{4x^2+4xy-1}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{(4x^2+4xy-1)[-(4y+4x\frac{dy}{dx})-8y\cdot \frac{dy}{dx}]-(1-4xy-4y^2)(8x+4y+4x\cdot \frac{dy}{dx})}{(4x^2+4xy-1{)}^2}$
at $x=0,y=1,\frac{dy}{dx}=3$
$\therefore \frac{d^{2}y}{d{x}^2}=\frac{(-1)[-4-8\times 3]-(1-4)\left(4\right)}{1}=28+12=40$