## Thomas' Calculus 13th Edition

a) Apply the chain rule as follows: $\dfrac{∂(2x^2-3y-4) }{∂x}= 4x+0+0=4x$ b) Apply the chain rule as follows: $\dfrac{∂(2x^2-3y-4) }{∂y}= 0-3+0=-3$