For \( x > y > 0 \), if \( x^5 y^6 = (x + y)^{11} \), then \( \frac{d^2y}{dx^2} \) is

0

The correct answer is Option (4) → 0

Given: $x^5 y^6 = (x+y)^{11},\ x>y>0$

Take logarithm: $\ln(x^5 y^6)=\ln\big((x+y)^{11}\big)\Rightarrow 5\ln x+6\ln y=11\ln(x+y)$

Differentiate w.r.t. $x$:

$\displaystyle \frac{5}{x}+\frac{6}{y}\frac{dy}{dx}= \frac{11}{x+y}\Big(1+\frac{dy}{dx}\Big)$

Rearrange terms containing $\frac{dy}{dx}$:

$\displaystyle \frac{6}{y}\frac{dy}{dx}-\frac{11}{x+y}\frac{dy}{dx}=\frac{11}{x+y}-\frac{5}{x}$

Factor $\frac{dy}{dx}$ and simplify both sides (common denominators):

$\displaystyle \frac{dy}{dx}\Big(\frac{6}{y}-\frac{11}{x+y}\Big)=\frac{11}{x+y}-\frac{5}{x}$

$\displaystyle \frac{dy}{dx}\cdot\frac{6x-5y}{y(x+y)}=\frac{6x-5y}{x(x+y)}$

Cancel $6x-5y$ (if $6x-5y=0$ the following relation still holds by continuity):

$\displaystyle \frac{dy}{dx}=\frac{y}{x}$

Differentiate again to get $\frac{d^2y}{dx^2}$:

$\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}\Big(\frac{y}{x}\Big)=\frac{x\frac{dy}{dx}-y}{x^2}$

Substitute $\frac{dy}{dx}=\frac{y}{x}$:

$\displaystyle \frac{d^2y}{dx^2}=\frac{x\cdot\frac{y}{x}-y}{x^2}=\frac{y-y}{x^2}=0$

Answer: \(\displaystyle \frac{d^2y}{dx^2}=0\)