Questions: f(x, y) = sin^(-1)(xy) (Answer: ∂f/∂x = y/√(1-x^2 y^2), ∂f/∂y = x/√(1-x^2 y^2))

Solution Steps

To find the partial derivatives of the function \( f(x, y) = \sin^{-1}(xy) \), we will use the chain rule for differentiation. Specifically, we need to differentiate \( f \) with respect to \( x \) while treating \( y \) as a constant, and then differentiate \( f \) with respect to \( y \) while treating \( x \) as a constant.

We start with the function defined as: \[ f(x, y) = \sin^{-1}(xy) \]

Using the chain rule, the partial derivative of \( f \) with respect to \( x \) is given by: \[ \frac{\partial f}{\partial x} = \frac{y}{\sqrt{1 - (xy)^2}} = \frac{y}{\sqrt{-x^2y^2 + 1}} \]

Similarly, the partial derivative of \( f \) with respect to \( y \) is: \[ \frac{\partial f}{\partial y} = \frac{x}{\sqrt{1 - (xy)^2}} = \frac{x}{\sqrt{-x^2y^2 + 1}} \]

Final Answer