## University Calculus: Early Transcendentals (3rd Edition)

The slope of the curve at $(1,0)$ is $-1$ and the slope of the curve at $(1,-1)$ is $1$.
$$(x^2+y^2)^2=(x-y)^2$$ 1) Find $dy/dx$: Differentiate both sides of the equation with respect to $x$: $$2(x^2+y^2)\frac{d}{dx}(x^2+y^2)=2(x-y)\frac{d}{dx}(x-y)$$ $$(x^2+y^2)\Big(2x+2y\frac{dy}{dx}\Big)=(x-y)\Big(1-\frac{dy}{dx}\Big)$$ $$(2x^3+2xy^2)+(2x^2y+2y^3)\frac{dy}{dx}=(x-y)-(x-y)\frac{dy}{dx}$$ Isolate the terms with $dy/dx$ into one side and solve for $dy/dx$: $$(2x^2y+2y^3)\frac{dy}{dx}+(x-y)\frac{dy}{dx}=x-y-2x^3-2xy^2$$ $$(x-y+2x^2y+2y^3)\frac{dy}{dx}=x-y-2x^3-2xy^2$$ $$\frac{dy}{dx}=\frac{x-y-2x^3-2xy^2}{x-y+2x^2y+2y^3}$$ 2) The slope of the curve at a point $(a,b)$ is the value of the derivative $dy/dx$ at that point. - At $(1,0)$: $$\frac{dy}{dx}=\frac{1-0-2\times1^3-2\times1\times0^2}{1-0+2\times1^2\times0+2\times0^3}=\frac{1-2}{1}=-1$$ The slope of the curve at $(1,0)$ is $-1$. - At $(1,-1)$: $$\frac{dy}{dx}=\frac{1-(-1)-2\times1^3-2\times1\times(-1)^2}{1-(-1)+2\times1^2\times(-1)+2\times(-1)^3}=\frac{1+1-2-2}{1+1-2-2}=1$$ The slope of the curve at $(1,-1)$ is $1$.