Question

Find $\frac{dy}{dx}$in the following questions:

$x^2+x^{2}y+x{y}^2+y^3=81.$

Answer 1

Given, $x^2+x^{2}y+x{y}^2+y^3=81.$

Differentiating both sides w.r.t. x, we get

$\frac{d}{dx}{x}^2+x^{2}y+x{y}^2+y^3=81$

$3x^2+\frac{d}{dx}\left(x^{2}y\right)+\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(x{y}^2\right)+3y^2\frac{dy}{dx}=0$

$\Rightarrow 3x^2+x^2\frac{dy}{dx}+y\left(2x\right)+x\left(2y\frac{dy}{dx}\right)+y^2.1+3y^2\frac{dy}{dx}=0\left(Usingproductrule\frac{d}{dx}\right(u.v)=u\frac{d}{dx}v+v\frac{d}{dx}u)$

$\Rightarrow (x^2+2xy+3y^2)\frac{dy}{dx}=-3x^2-2xy-y^2\Rightarrow \frac{dy}{dx}=-\frac{(3x^2+2xy+y^2)}{x^2+2xy+3y^2}$