F(x,y) = e^(2x+y)
Fx = 2e^(2x+y)
Fy = e^(2x+y)
Fx(1,-1) = 2e^(2-1) = 2e
Fy(1,-1) = e^(2-1) = e
F(x, y, z) = xy + xz -yz
Fx = y+z
Fy = x-z
Fz = x-y
Fx(0,-1,1) = 0
Fy(0,-1,1) = -1
Fz(0,-1,1) = 1
F(x,y,z) = 2x^0.2 y^0.8 + z^2
Fx = 0.4x^-0.8 y^0.8
Fy = 1.6x^0.2 y^-0.2
Fz = 2z
Fx(0,-1,1) = undefined
Fy(0,-1,1) = 0
Fz(0,-1,1) = 2
I think I have the right answer, but I am not 100% sure how to do the values at (1,-1) and the one below at (0,-1, 1). Please explain how I need to approach how to do these. Thank You!
Calculate the partial derivative @f/@x, @f/@y and
@f/@x | (1,-1), and @f/@y | (1,-1)
f(x,y) = e^(2x+y)
I got f(y) = 0 and f(x) = e^xy
Find @f/@x, @f/@y, @f/@z and their
values at (0,-1, 1)
1.) f(x, y, z) = xy + xz -yz
I got
f(x) = 0
f(y) = 0
f(z) = 0
2.) f(x, y, z) = 2x^0.2 y^0.8 + z^2
Here is what I got.
f(x) = 0.4(y^0.8) / x^0.8
f(y) = 1.6x^0.2 / y^0.2
f(z) = 2z
Do I substitute on the (1,-1) and do the same on the (0,-1,1) to find that point on the graph.
1 answer
1 answer