Question

If $(x^2+xy+3y^2)=1$, what will be the value of ${\left(x+6y\right)}^3\cdot \frac{d^{2}y}{d{x}^2}$ ?

Answer 1

We have,

$x^2+xy+3y^2=1$ …….. (1)

On differentiating both sides w.r.t $x$, we get

$2x+y+x\frac{dy}{dx}+6y\frac{dy}{dx}=0$

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

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

On differentiating both sides w.r.t $x$, we get

$\frac{d^{2}y}{d{x}^2}=\frac{d(-\frac{(2x+y)}{(x+6y)})}{dx}$

$\frac{d^{2}y}{d{x}^2}=-\frac{[\left(\frac{d(2x+y)}{dx}\right)(x+6y)-(2x+y)\left(\frac{d(x+6y)}{dx}\right)]}{{(x+6y)}^2}$ $[\because \frac{d\left(\frac{u}{v}\right)}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}]$

$\frac{d^{2}y}{d{x}^2}=-\frac{[(2+\frac{dy}{dx})(x+6y)-(2x+y)(1+6\frac{dy}{dx})]}{{(x+6y)}^2}$

$\frac{d^{2}y}{d{x}^2}=-\frac{[2x+12y+(x+6y)\frac{dy}{dx}-(2x+12x\frac{dy}{dx}+y+6y\frac{dy}{dx})]}{{(x+6y)}^2}$

$\frac{d^{2}y}{d{x}^2}=-\frac{[2x+12y+(x+6y)\frac{dy}{dx}-(2x+y+(6y+12x)\frac{dy}{dx})]}{{(x+6y)}^2}$

$\frac{d^{2}y}{d{x}^2}=-\frac{[2x+12y+(x+6y)\frac{dy}{dx}-2x-y-6(2x+y)\frac{dy}{dx}]}{{(x+6y)}^2}$

$\frac{d^{2}y}{d{x}^2}=-\frac{[2x+12y-2x-y+(x+6y-12x-6y)\frac{dy}{dx}]}{{(x+6y)}^2}$

$\frac{d^{2}y}{d{x}^2}=-\frac{(11y-11x\frac{dy}{dx})}{{(x+6y)}^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{11(x\frac{dy}{dx}-y)}{{(x+6y)}^2}$

From equation (2),

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

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

$\frac{d^{2}y}{d{x}^2}=-\frac{11(x(2x+y)+y(x+6y))}{{(x+6y)}^3}$

${(x+6y)}^3\frac{d^{2}y}{d{x}^2}=-11(2x^2+xy+xy+6y^2)$

${(x+6y)}^3\frac{d^{2}y}{d{x}^2}=-11(2x^2+2xy+6y^2)$

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

From equation (1),

${(x+6y)}^3\frac{d^{2}y}{d{x}^2}=-22\times 1$

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

Hence, this is the answer.