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

Step 1: Strip off the inverse sine.
Let \(z=\sin^2 u=\left(\dfrac{x^{1/3}+y^{1/3}}{x^{1/2}-y^{1/2}}\right)\). So \(z\) is the homogeneous expression and \(u\) is tucked inside through \(z=\sin^2 u\).

Step 2: Find the degree of homogeneity of z.
Top terms have degree \(\tfrac13\); bottom terms have degree \(\tfrac12\). The whole ratio is homogeneous of degree \(n=\tfrac13-\tfrac12=-\tfrac16\).

Step 3: Apply Euler's theorem to z.
\[x z_x+y z_y=n z=-\tfrac16 z.\]
Step 4: Relate z to u.
Since \(z=\sin^2 u\), we get \(z_x=2\sin u\cos u\,u_x\) and similarly for y. So \[x z_x+y z_y=2\sin u\cos u\,(x u_x+y u_y).\]
Step 5: Combine the two expressions.
\[2\sin u\cos u\,(x u_x+y u_y)=-\tfrac16\sin^2 u.\]
Step 6: Solve for the required sum.
Divide both sides by \(2\sin u\cos u\): \[x u_x+y u_y=-\frac{1}{12}\,\frac{\sin u}{\cos u}=-\frac{1}{12}\tan u.\] This is option 1.
\[ \boxed{x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=-\tfrac{1}{12}\tan u} \]