Question

U= Sin^-1[y\x]
Verify Uxy=Uyx​

Answer 1

Answer:

good evening.........

Answer 2

Answer:

We've successfully verified the given equation, along with all the required steps.

Step-by-step explanation:

To verify that Uxy = Uyx, we first need to find the partial derivatives of U with respect to x and y.

Starting with U = sin⁻¹(y/x), we can use the chain rule to find the partial derivative with respect to x:

∂U/∂x = (1/√(1-(y/x)²)) * (-y/x²) = -y/[(x²(1-(y/x)²))]

Similarly, the partial derivative with respect to y can be found using the chain rule:

∂U/∂y = (1/√(1-(y/x)²)) * (1/x) = 1/[(x√(1-(y/x)²))]

Now we can check if Uxy = Uyx:

Uxy = -y/[(x²(1-(y/x)²))]

Uyx = 1/[(x√(1-(y/x)²))]

To simplify the expression, we can use the

identity sin²θ + cos²θ = 1, which implies 1 - sin²θ = cos²θ.

Substituting sin⁻¹(y/x) for θ, we get cos²(sin⁻¹(y/x)) = 1 - (y/x)².

Using this identity, we can simplify Uxy and Uyx:

Uxy = -y/[(x²cos²(sin⁻¹(y/x))] = -y/[(x²(1-(y/x)²))]

Uyx = 1/[(xsin(sin⁻¹(y/x))] = 1/[(x√(1-(y/x)²))]

As we can see, Uxy = Uyx, which means that the mixed partial derivatives of U exist and are equal.

This result is not surprising, since sin⁻¹(y/x) is a well-behaved function with continuous second partial derivatives in its domain. However, the computation highlights the importance of checking for mixed partial derivatives, especially when dealing with more complex functions, as the order of differentiation can affect the outcome.

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