Applying Euler's Theorem to Find Partial Derivatives

The problem asks us to find the value of \(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\) for the function \(u=\sin^{-1}\frac{x+2y}{x^8+y^8}\).

This expression \(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\) strongly suggests the application of Euler's theorem for homogeneous functions.

Let's first examine the function \(u\). It is given as \(u = \sin^{-1}\left(\frac{x+2y}{x^8+y^8}\right)\). This function \(u\) itself is not directly a homogeneous function.

However, consider the term inside the inverse sine function. Let \(v = \sin(u)\). So, \(v = \frac{x+2y}{x^8+y^8}\).

Let's check if this function \(v(x, y)\) is a homogeneous function. A function \(f(x, y)\) is homogeneous of degree \(n\) if \(f(tx, ty) = t^n f(x, y)\) for some constant \(n\).

Let's substitute \(tx\) for \(x\) and \(ty\) for \(y\) in \(v(x, y)\):

\[v(tx, ty) = \frac{(tx) + 2(ty)}{(tx)^8 + (ty)^8}\] \[v(tx, ty) = \frac{t(x + 2y)}{t^8x^8 + t^8y^8}\] \[v(tx, ty) = \frac{t(x + 2y)}{t^8(x^8 + y^8)}\] \[v(tx, ty) = t^{1-8} \frac{x + 2y}{x^8 + y^8}\] \[v(tx, ty) = t^{-7} \frac{x + 2y}{x^8 + y^8}\] \[v(tx, ty) = t^{-7} v(x, y)\]

Since \(v(tx, ty) = t^{-7} v(x, y)\), the function \(v(x, y) = \sin(u)\) is a homogeneous function of degree \(n = -7\).

According to Euler's theorem for homogeneous functions, if \(v(x, y)\) is a homogeneous function of degree \(n\), then:

\[x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y} = n v\]

In our case, \(v = \sin(u)\) and \(n = -7\). So, applying Euler's theorem to \(v(x, y) = \sin(u)\):

\[x \frac{\partial (\sin u)}{\partial x} + y \frac{\partial (\sin u)}{\partial y} = -7 (\sin u)\]

Now, we need to compute the partial derivatives of \(\sin u\) with respect to \(x\) and \(y\). Using the chain rule, we have:

\[\frac{\partial (\sin u)}{\partial x} = \cos u \frac{\partial u}{\partial x}\] \[\frac{\partial (\sin u)}{\partial y} = \cos u \frac{\partial u}{\partial y}\]

Substitute these expressions back into the equation from Euler's theorem:

\[x \left(\cos u \frac{\partial u}{\partial x}\right) + y \left(\cos u \frac{\partial u}{\partial y}\right) = -7 \sin u\]

Factor out \(\cos u\) from the left side:

\[\cos u \left(x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}\right) = -7 \sin u\]

Assuming \(\cos u \neq 0\), we can divide both sides by \(\cos u\) to solve for \(x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}\):

\[x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{-7 \sin u}{\cos u}\]

Since \(\frac{\sin u}{\cos u} = \tan u\), the expression simplifies to:

\[x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = -7 \tan u\]

This matches one of the given options.

Revision Table: Key Concepts

Additional Information on Euler's Theorem

Euler's theorem provides a powerful shortcut for certain types of partial derivative problems involving homogeneous functions. Here are some additional points:

• Conditions: The theorem requires the function to be homogeneous and have continuous first-order partial derivatives in the domain being considered.
• Generalization: For a homogeneous function of \(m\) variables \(f(x_1, x_2, \dots, x_m)\) of degree \(n\), the theorem states: \[\sum_{i=1}^{m} x_i \frac{\partial f}{\partial x_i} = n f\]
• Identifying Homogeneity: A quick way to check for homogeneity in rational functions (like \(\frac{P(x,y)}{Q(x,y)}\)) is to find the degree of each term in the numerator and denominator. If every term in the numerator has the same degree \(d_1\) and every term in the denominator has the same degree \(d_2\), then the function is homogeneous of degree \(n = d_1 - d_2\). In our case, the numerator \(x+2y\) has terms of degree 1. The denominator \(x^8+y^8\) has terms of degree 8. So the degree is \(1-8 = -7\).
• Inverse Functions: When dealing with inverse trigonometric or inverse hyperbolic functions like \(u = f^{-1}(v)\), if \(v\) is homogeneous, Euler's theorem is applied to the outer function \(v = f(u)\), and then the chain rule is used to bring in the partial derivatives of \(u\), as demonstrated in this problem.