## Thomas' Calculus 13th Edition

$h_x=2πcos(2πx+y-3z)$ $h_y=cos(2πx+y-3z)$ $h_z=-3cos(2πx+y-3z)$
Take the first partial derivatives of the given function. When taking partial derivative with respect to x, treat y and z as constants. Use the same method when taking partial derivative with respect to y and z: $h_x=cos(2πx+y-3z)\times(2π+0-0)=2πcos(2πx+y-3z)$ $h_y=cos(2πx+y-3z)\times(0+1-0)=cos(2πx+y-3z)$ $h_z=cos(2πx+y-3z)\times(0+0-3)=-3cos(2πx+y-3z)$