Question

Suppose y'=f(x, y) = xy/cos(x). a. ∂f/∂y = ________________ . b.Since the function f(x, y) is __________________ at the point (0, 0), the partial derivative ∂f/∂y ______________ and is ___________ at and near the point (0, 0), the solution to y prime = f(x, y) ________ near y(0) = 0

Answer 1

The partial derivative ∂f/∂y of f(x, y) = xy/cos(x) is x/cos(x). The function f(x, y) is continuous at and near the point (0, 0). The solution to y' = f(x, y) near y(0) = 0 depends on the initial conditions.

Step-by-step explanation:

a. To find the partial derivative ∂f/∂y, we differentiate f(x, y) with respect to y while treating x as a constant. In this case, since f(x, y) = xy/cos(x), the derivative of xy/cos(x) with respect to y is just x/cos(x), since the derivative of y with respect to y is 1 and all other terms are treated as constants.

b. Since the function f(x, y) = xy/cos(x) is continuous at the point (0, 0), the partial derivative ∂f/∂y is defined and continuous at and near the point (0, 0). The solution to y' = f(x, y) near y(0) = 0 depends on the specific initial conditions and cannot be determined without further information.