How do i do this any help please

f(x, y , z) = xy^2 + 2yz^3

evaluate ∂.f / ∂.x = at (x , y , z) = (6, -1 , 1).................


evaluate ∂.f / ∂.y= at (x , y , z) = (6, -1 , 1).................



evaluate ∂.f / ∂.z = at (x , y , z) = (6, -1 , 1).................

thank you

For partial differentiation, treat two of the variables as constants when differentiating with respect to the third; so for example, when differentiating partially with respect to x, treat y and z as constants,

i) ∂f/∂x = y² + 0
At (6, -1, 1) we get ∂f/∂x = 1

ii) ∂f/∂y = 2xy + 2z³
At (6, -1, 1) we get ∂f/∂y = -12 + 2 = -10

iii) ∂f/∂z = 0 + 6yz²
At (6, -1, 1) we get ∂f/∂z = -6

1)
Treat y and z as co-efficients:
∂f/∂x = y^2 + 0 = y^2

At (6, -1, 1)
∂f/∂x = (-1)^2 = 1


2)
Treat x and z as co-efficients:
∂f/∂y = xy/2 + 2z^3

At (6, -1, 1)
∂f/∂y = 6*(-1)/2 + 2*1^3 = -3 + 2 = -1


3)
Treat x and y as co-efficients:
∂f/∂z = 0 + (2yz^2)/3 = (2yz^2)/3

At (6, -1, 1)
∂f/∂z = (2*(-1)*1^2)/3 = -2/3

Hope this helps!

When determining a partial derivative, treat all other variables as if they were constants.

∂.f / ∂.x = (1 * x^0)y^2 + 0 = y^2

∂.f(6, -1, 1) / ∂.x = (-1)^2 = 1