Final answer:
To find all second partial derivatives of the function f(x,y) = e⁴ʳ - x sin(y²) + y³, we differentiate twice with respect to both x and y and calculate the mixed partial derivatives, yielding (x,y), fyy(x,y), and fxy(x,y) = fyx(x,y).
Step-by-step explanation:
The question requires finding all second partial derivatives of the function f(x,y) = e⁴ʳ - x sin(y²) + y³. First, we find the first partial derivatives with respect to x and y, and then differentiate these results a second time to obtain the second partial derivatives.
The first partial derivative of f with respect to x is given by fx(x,y) = 4e⁴ʳ - sin(y²) and the second partial derivative with respect to x is (x,y) = 16e⁴ʳ.
The first partial derivative of f with respect to y is fy(x,y) = -2xy cos(y²) + 3y² and the second partial derivatives with respect to y are fyy(x,y) = -2x(2y cos(y²) + y⁴ sin(y²)) + 6y.
The mixed partial derivatives are fxy(x,y) = fyx(x,y) = -2y cos(y²).