Find the four second partial derivatives and evaluate each at the given point.
Function
f(x, y) = x^3 + 2xy^3 − 9y
Point
(9, 2)
fxx(9, 2) =
fxy(9, 2) =
fyx(9, 2) =
fyy(9, 2) =
Fx = 3x^2 + 2y^3
Fy = 6xy^2 - 9
Fxx = 6x
Fxy = 6y^2 = Fyx
Fyy = 12x
Now just plug in the numbers
do this just like the one above
for example
df/dx = 3 x^2 + 2 y^3
df/dy = 6 x y^2 - 9
d/dx (df/dx) = 6 x
etc
To find the four second partial derivatives of the function f(x, y) = x^3 + 2xy^3 − 9y at the point (9, 2), we need to calculate the partial derivatives with respect to x and y twice.
First, let's find the first partial derivatives:
fx(x, y) = d/dx(x^3 + 2xy^3 − 9y) = 3x^2 + 2y^3
fy(x, y) = d/dy(x^3 + 2xy^3 − 9y) = 6xy^2 - 9
Now, let's find the second partial derivatives:
fxx(x, y) = d/dx(3x^2 + 2y^3) = 6x
fxy(x, y) = d/dy(3x^2 + 2y^3) = 6xy^2
fyx(x, y) = d/dx(6xy^2 - 9) = 6y^2
fyy(x, y) = d/dy(6xy^2 - 9) = 12xy
Now, let's evaluate each of these derivatives at the given point (9, 2):
fxx(9, 2) = 6(9) = 54
fxy(9, 2) = 6(9)(2^2) = 6(9)(4) = 216
fyx(9, 2) = 6(2^2) = 6(4) = 24
fyy(9, 2) = 12(9)(2) = 12(18) = 216
Therefore, we have:
fxx(9, 2) = 54
fxy(9, 2) = 216
fyx(9, 2) = 24
fyy(9, 2) = 216
To find the second partial derivatives of the function f(x, y) = x^3 + 2xy^3 - 9y, we need to take the partial derivative of the function twice with respect to each variable.
The first partial derivative with respect to x (f_x) is obtained by differentiating each term with respect to x while treating y as a constant. Similarly, the first partial derivative with respect to y (f_y) is obtained by differentiating each term with respect to y while treating x as a constant.
Let's find the first partial derivatives:
f_x = d/dx (x^3) + d/dx (2xy^3) - d/dx (9y)
= 3x^2 + 2y^3 * d/dx (x) + 0 - 0
= 3x^2 + 2xy^3
f_y = d/dy (x^3) + d/dy (2xy^3) - 9 * d/dy (y)
= 0 + 3y^2 * d/dy (x) + 2x * d/dy (y^3) - 9 * 1
= 2x * 3y^2 + 0 - 9
= 6xy^2 - 9
Now, to find the second partial derivatives, we need to take the partial derivatives of f_x and f_y with respect to x and y, respectively.
The partial derivative f_xx (second partial derivative with respect to x) is obtained by differentiating f_x with respect to x, while treating y as a constant:
f_xx = d/dx (3x^2 + 2xy^3)
= 6x + 2y^3 * d/dx (x)
= 6x + 2xy^3
The partial derivative f_xy (second partial derivative with respect to x and y) is obtained by differentiating f_x with respect to y, while treating x as a constant:
f_xy = d/dy (3x^2 + 2xy^3)
= 0 + 2y^3 * d/dy (x)
= 2y^3 * 0
= 0
The partial derivative f_yx (second partial derivative with respect to y and x) is obtained by differentiating f_y with respect to x, while treating y as a constant:
f_yx = d/dx (6xy^2 - 9)
= 6y^2 * d/dx (x) + 0
= 6y^2
The partial derivative f_yy (second partial derivative with respect to y) is obtained by differentiating f_y with respect to y, while treating x as a constant:
f_yy = d/dy (6xy^2 - 9)
= 0 + 2y * 6x * d/dy (y) - 0
= 12xy
Now, we can evaluate each of these second partial derivatives at the given point (9, 2):
f_xx(9, 2) = 6(9) + 2(2)^3 = 54 + 16 = 70
f_xy(9, 2) = 0
f_yx(9, 2) = 6(2)^2 = 6(4) = 24
f_yy(9, 2) = 12(9)(2) = 216
Therefore,
fxx(9, 2) = 70
fxy(9, 2) = 0
fyx(9, 2) = 24
fyy(9, 2) = 216