Answer to Question #194256 in Calculus for Maureen

Question

Use implicit differenciation to obtain dy over dx

X over y+1 = xsquared+3y

Answer 1

Given $\frac{x}{y}+1 = x^2 + 3y$

Differentiating both sides,

$\frac{d}{dx}( \frac{x}{y}+1) =\frac{d}{dx}( x^2 + 3y)$

$\frac{1}{y} - \frac{x}{y^2}\frac{dy}{dx} = 2x+3\frac{dy}{dx}$

It can be written as,

$\frac{1}{y} - 2x= 3\frac{dy}{dx}+\frac{x}{y^2}\frac{dy}{dx}$

$\frac{1-2xy}{y}= (3+\frac{x}{y^2})\frac{dy}{dx}$

$\frac{1-2xy}{y} = \frac{3y^2+x}{y^2}\frac{dy}{dx}$

$\frac{dy}{dx} = \frac{(1-2xy)y}{3y^2+x}$