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

- At $(-3,2)$, the slope of the curve is $-27/8$. - At $(-3,-2)$, the slope of the curve is $27/8$. - At $(3,-2)$, the slope of the curve is $-27/8$. - At $(3,2)$, the slope of the curve is $27/8$.
$$y^4-4y^2=x^4-9x^2$$ 1) Find the derivative of the function using implicit differentiation: $$4y^3y'-8yy'=4x^3-18x$$ $$2y^3y'-4yy'=2x^3-9x$$ $$y'(2y^3-4y)=2x^3-9x$$ $$y'=\frac{2x^3-9x}{2y^3-4y}$$ 2) The slope of the curve at $(-3,2)$ is $$y'=\frac{2\times(-3)^3-9(-3)}{2\times2^3-4\times2}=\frac{2\times(-27)+27}{16-8}=\frac{-27}{8}=-\frac{27}{8}$$ - The slope of the curve at $(-3,-2)$ is $$y'=\frac{2\times(-3)^3-9(-3)}{2\times(-2)^3-4\times(-2)}=\frac{2\times(-27)+27}{-16+8}=\frac{-27}{-8}=\frac{27}{8}$$ - The slope of the curve at $(3,-2)$ is $$y'=\frac{2\times3^3-9\times3}{2\times(-2)^3-4\times(-2)}=\frac{2\times27-27}{-16+8}=\frac{27}{-8}=-\frac{27}{8}$$ - The slope of the curve at $(3,2)$ is $$y'=\frac{2\times3^3-9\times3}{2\times2^3-4\times2}=\frac{2\times27-27}{16-8}=\frac{27}{8}$$