Question

mes%20y%20%5Ctext%20%7B.%20Find%20%7D%20%5Cfrac%7Bd%5E%7B2%7D%20y%7D%7Bd%20x%5E%7B2%7D%7D%20%5Ctext%20%7B%20at%20%7D%20x%3D0%20%5Ctext%20%7B.%20%7D%5Cend%7Bequation%7D" id="TexFormula1" title="\begin{equation}\text { Question: If } \ln (x+y)=4 \times y \text {. Find } \frac{d^{2} y}{d x^{2}} \text { at } x=0 \text {. }\end{equation}" alt="\begin{equation}\text { Question: If } \ln (x+y)=4 \times y \text {. Find } \frac{d^{2} y}{d x^{2}} \text { at } x=0 \text {. }\end{equation}" align="absmiddle" class="latex-formula">

Answer 1

I think you meant to say

$\ln(x+y) = 4xy$

and not "4 times y" on the right side (which would lead to a complex value for y when x = 0). Note that when x = 0, the equation reduces to ln(y) = 0, so that y = 1.

Implicitly differentiating both sides with respect to x, taking y = y(x), and solving for dy/dx gives

$\dfrac{1+\frac{dy}{dx}}{x+y} = 4y + 4x\dfrac{dy}{dx}$

$\implies \dfrac{dy}{dx} = \dfrac{4xy+4y^2-1}{1-4x^2-4xy}$

Note that when x = 0 and y = 1, we have dy/dx = 3.

Differentiate both sides again with respect to x :

$\dfrac{d^2y}{dx^2} = \dfrac{(1-4x^2-4xy)\left(4y+4x\frac{dy}{dx}+8y\frac{dy}{dx}\right)-(4xy+4y^2-1)\left(-8x-4y-4x\frac{dy}{dx}\right)}{(1-4x^2-4xy)^2}$

No need to simplify; just plug in x = 0, y = 1, and dy/dx = 3 to get

$\dfrac{d^2y}{dx^2} \bigg|_{x=0} = \boxed{40}$