Question

If \(u=\sin^{-1}\left(\dfrac{x^{1/3}+y^{1/3}}{x^{1/2}-y^{1/2}}\right)^{1/2}\), then \(x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=\)

• A. \(-\dfrac{1}{12}\tan u\)
• B. \(\dfrac{1}{12}\tan u\)
• C. \(\dfrac{1}{6}\cos u\)
• D. \(-\dfrac{1}{6}\sin u\)

Hint:

This uses Euler's theorem for homogeneous functions. If a function \(z\) is homogeneous of degree \(n\), then \(x\,z_x+y\,z_y=n\,z\).

Answer 1

The Correct Option is A

Concept:
This uses Euler's theorem for homogeneous functions. If a function \(z\) is homogeneous of degree \(n\), then \(x\,z_x+y\,z_y=n\,z\). When the function is wrapped inside an inverse sine, we put \(z=\sin u\) and carry the degree through.

Step 1:
Let \(z=\left(\dfrac{x^{1/3}+y^{1/3}}{x^{1/2}-y^{1/2}}\right)^{1/2}\), so \(u=\sin^{-1}z\), i.e. \(z=\sin u\).

Step 2:
Find the degree of \(z\). The top \(x^{1/3}+y^{1/3}\) is homogeneous of degree \(\tfrac13\); the bottom \(x^{1/2}-y^{1/2}\) is degree \(\tfrac12\). So the ratio has degree \(\tfrac13-\tfrac12=-\tfrac16\). Raising it to the power \(\tfrac12\) multiplies the degree by \(\tfrac12\): \[n=-\tfrac16\times\tfrac12=-\tfrac{1}{12}.\]

Step 3:
Apply Euler's theorem to \(z=\sin u\): \[x\,z_x+y\,z_y=n\,z=-\tfrac{1}{12}\sin u.\]

Step 4:
Since \(z=\sin u\), the chain rule gives \(z_x=\cos u\,u_x\) and \(z_y=\cos u\,u_y\). Substituting: \[\cos u\left(x\,u_x+y\,u_y\right)=-\tfrac{1}{12}\sin u.\] Divide by \(\cos u\): \[x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=-\tfrac{1}{12}\tan u.\]

Answer: Option (1) — \(-\dfrac{1}{12}\tan u\).