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
To calculate the partial derivatives of the function f(x,y) = e^(2x+y), you can use the chain rule.
1. The partial derivative @f/@x represents the rate of change of f with respect to x, while keeping y constant. To calculate it, we differentiate f(x,y) with respect to x:
@f/@x = d/dx (e^(2x+y))
= (d/dx) (e^(2x+y))
= 2e^(2x+y)
2. The partial derivative @f/@y represents the rate of change of f with respect to y, while keeping x constant. To calculate it, we differentiate f(x,y) with respect to y:
@f/@y = d/dy (e^(2x+y))
= (d/dy) (e^(2x+y))
= e^(2x+y)
To find the values of @f/@x and @f/@y at the point (1,-1), substitute x=1 and y=-1 into the partial derivatives you calculated above:
@f/@x | (1,-1) = 2e^(2(1)+(-1)) = 2e^1 = 2e
@f/@y | (1,-1) = e^(2(1)+(-1)) = e^1 = e
For the function f(x, y, z) = xy + xz - yz, to find the partial derivatives @f/@x, @f/@y, and @f/@z, you can differentiate each term with respect to the corresponding variable and sum them up:
1. @f/@x = d/dx (xy + xz - yz)
= y + z - 0 (since the derivative of a constant term is 0)
= y + z
2. @f/@y = d/dy (xy + xz - yz)
= x - z - z
= x - 2z
3. @f/@z = d/dz (xy + xz - yz)
= x - y - y
= x - 2y
To find the values of @f/@x, @f/@y, and @f/@z at the point (0,-1,1), substitute x=0, y=-1, and z=1 into the partial derivatives you calculated above:
@f/@x | (0,-1,1) = -1 + 1 = 0
@f/@y | (0,-1,1) = 0 - 2(1) = -2
@f/@z | (0,-1,1) = 0 - 2(-1) = 2
For the function f(x, y, z) = 2x^0.2 y^0.8 + z^2, we can take the partial derivatives as follows:
1. @f/@x = d/dx (2x^0.2 y^0.8 + z^2)
= 0.4x^(-0.8) y^0.8 + 0
= 0.4(y^0.8) / x^0.8
2. @f/@y = d/dy (2x^0.2 y^0.8 + z^2)
= 1.6x^0.2 y^(-0.2) + 0
= 1.6x^0.2 / y^0.2
3. @f/@z = d/dz (2x^0.2 y^0.8 + z^2)
= 0 + 2z
= 2z
To find the values of @f/@x, @f/@y, and @f/@z at the point (0,-1,1), substitute x=0, y=-1, and z=1 into the partial derivatives you calculated above:
@f/@x | (0,-1,1) = 0.4(-1^0.8) / 0^0.8 = 0
@f/@y | (0,-1,1) = 1.6(0^0.2) / (-1)^0.2 = 0
@f/@z | (0,-1,1) = 2(1) = 2