implicit differentiation1.if y^2=× 2.if ×^2+3×y+2y^2=3,find dy/dx at point (-1,2)

$x^2+3xy+2y^2=3$

Differentiate both sides of the equation.

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

Differentiate the left side of the equation.

$3x{y}^{\prime }+4y{y}^{\prime }+2x+3y$

Since $3$ is constant with respect to $x$, the derivative of $3$ with respect to $x$ is $0$.

Reform the equation by setting the left side equal to the right side.

$3x{y}^{\prime }+4y{y}^{\prime }+2x+3y=0$

Solve for $y^{\prime }$

$y^{\prime }=-\frac{2x+3y}{3x+4y}$

Replace $y^{\prime }$ with $\frac{dy}{dx}$

$\frac{dy}{dx}=-\frac{2x+3y}{3x+4y}$

Put $(-1,2)$, where $x=-1,y=2$

$\frac{dy}{dx}=-\frac{2(-1)+3(2)}{3(-1)+4(2)}$

$\frac{dy}{dx}=-\frac{4}{5}$