Question

If $x^2+xy+3y^2=1$ then $(x+6y{)}^3\frac{d^{2}y}{d{x}^2}$ is

• A. $0$
• B. $-12$
• C. $22$
• D. $-22$

Answer 1

The correct option is D $-22$

Taking derivative wrt x,

$2x+(\frac{xdy}{dx}+y)+6y.\frac{\partial y}{\partial x}=0$

$\Rightarrow \frac{dy}{dx}(x+6y)=-2x-y$

$\Rightarrow \frac{dy}{dx}=\frac{-2x-y}{x+6y}$

Again differentiating,

$\frac{{\partial }^{2}y}{\partial x^2}=\frac{\left(\frac{-2-dy}{dx}\right)(x+6y)-(-2x-y)(1+6y)}{(x+6y{)}^2}$

$=\frac{y^{\prime }(-x-6y+12x+6y)-2x-12y+2x+y}{(x+6y{)}^2}$

$=\frac{y^{\prime }\left(11x\right)-11y}{(x+6y{)}^2}=\frac{11(x{y}^{\prime }-y)}{(x+6y{)}^2}$

$\Rightarrow y^{\prime \prime }(x+6y{)}^2=\frac{11(-2x^2-xy-xy-6y^2)}{(x+6y)}$

$y^{\prime \prime }(x+6y{)}^3=-22(x^2+xy+3y^2)=-22$