Let P(x,y)=5x^3y+8xy^3. Then Px(-1,1) is? And Pxy(-1,1) is?
To find Px(-1,1), we need to take the partial derivative of the function P(x, y) with respect to x and then evaluate it at the point (-1, 1).
To take the derivative, we differentiate each term of the function separately with respect to x, treating y as a constant.
P(x, y) = 5x^3y + 8xy^3
Now, let's compute the partial derivative with respect to x:
dP/dx = d(5x^3y)/dx + d(8xy^3)/dx.
To differentiate the first term, 5x^3y, we treat y as a constant and use the power rule for differentiation:
d(5x^3y)/dx = 5 * 3x^2y = 15x^2y.
Similarly, to differentiate the second term, 8xy^3, we treat y as a constant and differentiate x:
d(8xy^3)/dx = 8y^3.
Now, let's evaluate the partial derivative at (-1, 1):
Px(-1, 1) = 15(-1)^2(1) + 8(1)^3 = 15 - 8 = 7.
Therefore, Px(-1, 1) is equal to 7.
To find Pxy(-1, 1), we need to take the partial derivative of the function P(x, y) with respect to y, and then evaluate it at the point (-1, 1).
To take the derivative, we differentiate each term of the function separately with respect to y, treating x as a constant.
P(x, y) = 5x^3y + 8xy^3
Now, let's compute the partial derivative with respect to y:
dP/dy = d(5x^3y)/dy + d(8xy^3)/dy.
To differentiate the first term, 5x^3y, we treat x as a constant and use the power rule for differentiation:
d(5x^3y)/dy = 5x^3.
Similarly, to differentiate the second term, 8xy^3, we treat x as a constant and differentiate y:
d(8xy^3)/dy = 8 * 3xy^2 = 24xy^2.
Now, let's evaluate the partial derivative at (-1, 1):
Pxy(-1, 1) = 5(-1)^3 + 24(-1)(1)^2 = -5 - 24 = -29.
Therefore, Pxy(-1, 1) is equal to -29.