## Thomas' Calculus 13th Edition

a) $2x(y+2)$ b) $x^2-1$
a) Apply the chain rule as follows: $∂f/∂x = (x^2-1)'(y+2) + (x^2-1)(y+2)'= 2x (y +2) + (x^2-1)(0)= 2x(y+2)$ b) Apply the chain rule as follows: $∂f/∂y = (x^2-1)'(y+2) + (x^2-1)(y+2)'= 0(y+2) + (x^2-1)(1)= x^2-1$