Find the first partial derivative of the function.
f(x,y,z) = xy²z³ + 3yz
This is what i did:
f_x(x,y,z) = (1)(y²)(z³) +(0)(z³) + (0) + ((3y)(0))+ (0z)
= (y²)(z³)

f_y(x,y,z) = (0)(y²)(z³) +(2y)(z³) + (0) + ((3)(z)+ ((3y)(0))
=2yz³ + 3z

f_z(x,y,z) = (0)(y²)(z³) +(0)(z³) + 3z² + ((0)(2)+ ((3y)(1))
=3z² + 3y

correct answers:
f_x = y²z³
f_y =2xyz³ + 3z
f_z = 3xy²z² + 3y

Answer rating: 100% (QA)

fx y z xy 3yz When we take the partial derivative with respect to x we treat yand z as c... View the full answer