Chapter 14: Partial Derivatives - Section 14.3 - Partial Derivatives - Exercises 14.3 - Page 808: 62

(a) $3$ and (b) $-2$

(a) We know that the slope of the line tangent to a surface $f(x,y)$ at the point $(p,q)$ and lying in the plane $x=p$ is equal to $f_y(p,q)$. Thus, $f_y(x,y)=3y^2$ This means that $f_y(-1,1)=3 \cdot 1^2=3$ (b) We know that the slope of the line tangent to a surface $f(x,y)$ at the point $(p,q)$ and lying in the plane $y=q$ is equal to $f_x(p,q)$. Thus, $f_x(x,y)=2x$ This means that $f_x(-2,1)=2 (-1)=-2$