Question

Find the indicated partial derivative. f(x, y) = y sin−1(xy); fy 2, 1 4

Answer 1

The student is seeking the partial derivative with respect to y of the function f(x, y) evaluated at the point (2, 1/4). The derivative can be found using the chain rule and evaluating the expression at the given point.

Step-by-step explanation:

The student is asking for the partial derivative with respect to y of the function f(x, y) = y sin⁻¹(xy) evaluated at the point (2, 1/4). To find this, we first need to compute the partial derivative of f with respect to y, fy, and then substitute the given values of x and y into the resulting expression.

To compute fy, we apply the chain rule. Let u = xy. Then the partial derivative of f with respect to y is given by:

fy = sin⁻¹(u) + y * (d/dy)sin⁻¹(u)

Since u = xy, we have du/dy = x. And the derivative of sin⁻¹(u) with respect to u is 1/√(1-u²). Therefore,

fy = sin⁻¹(xy) + y * (x/√(1-(xy)²)).

Now, substituting x = 2 and y = 1/4 we get:

fy(2, 1/4) = sin⁻¹(1/2) + (1/4) * (2/√(1-(1/2)²))

This is the value of the partial derivative of f with respect to y at the point (2, 1/4).

Answer 2

The partial derivative of f(x, y) = y sin⁻¹(xy) with respect to y at the point (2, 1/4) is 1/4.

Step-by-step explanation:

Chain Rule: We'll use the chain rule to differentiate sin⁻¹(xy) with respect to y while treating x as a constant.

Inner Function: Let u = xy. Then, sin⁻¹(xy) becomes sin⁻¹(u). The derivative of sin⁻¹(u) with respect to u is 1/√(1 - u²).

Chain Rule Derivative: Applying the chain rule:

∂f/∂y = ∂(ysin⁻¹(xy))/∂y = y * ∂/∂y(sin⁻¹(u)) * ∂u/∂y

Inner Derivatives:

∂/∂y(sin⁻¹(u)) = 1/√(1 - u²)

∂u/∂y = x

Substitute and simplify:

∂f/∂y = y * 1/√(1 - (xy)²) * x = (xy) / √(1 - (xy)²)

Evaluate at the point:

∂f/∂y (2, 1/4) = (2 * (1/4)) / √(1 - (2 * 1/4)²) = 1/4

Therefore, the partial derivative of f with respect to y at (2, 1/4) is 1/4.