Question

x2+y3+2xy=4 find siop of tengent at point(1,1)

Answer 1

the slope of a curve at a point on the curve is equal to the derivative dy/dx at the point.

$x^2+y^3+2xy= 4\\\\differentiating\ wrt\ x:\\\\2x+3y^2\ y'+2x*y'+2y=0\\\\y'(3y^2+2x)=-2(x+y)\\\\y'=\frac{dy}{dx}=-2\frac{x+y}{3y^2+2x}\\\\substituting\ \ x=y=1\ \ at\ point\ (1,1),\ the\ slope\ is\\\\y'(1,1)=-2*\frac{1+1}{3+2}=-4/5$

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