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

The slope of the curve at $(\sqrt3/4,\sqrt3/2)$ is $-1$ and at $(\sqrt3/4,1/2)$ is $\sqrt3$.
$$y^4=y^2-x^2$$ 1) Find the derivative of the function using implicit differentiation: $$4y^3y'=2yy'-2x$$ $$2y^3y'=yy'-x$$ $$yy'-2y^3y'=x$$ $$y'(y-2y^3)=x$$ $$y'=\frac{x}{y-2y^3}$$ 2) The slope of the curve at $(\sqrt3/4,\sqrt3/2)$ is $$y'=\frac{\frac{\sqrt3}{4}}{\frac{\sqrt3}{2}-2\Big(\frac{\sqrt3}{2}\Big)^3}=\frac{\frac{\sqrt3}{4}}{\frac{\sqrt3}{2}-\frac{2\times\sqrt{27}}{8}}=\frac{\frac{\sqrt3}{4}}{\frac{\sqrt3}{2}-\frac{3\sqrt3}{4}}$$ $$y'=\frac{\frac{\sqrt3}{4}}{\frac{2\sqrt3-3\sqrt3}{4}}=\frac{\frac{\sqrt3}{4}}{\frac{-\sqrt3}{4}}=-1$$ - The slope of the curve at $(\sqrt3/4,1/2)$ is $$y'=\frac{\frac{\sqrt3}{4}}{\frac{1}{2}-2\Big(\frac{1}{2}\Big)^3}=\frac{\frac{\sqrt3}{4}}{\frac{1}{2}-\frac{2\times1}{8}}=\frac{\frac{\sqrt3}{4}}{\frac{1}{2}-\frac{1}{4}}$$ $$y'=\frac{\frac{\sqrt3}{4}}{\frac{1}{4}}=\sqrt3$$