Question

If \(u=\dfrac{xy}{x+y}\), then \(x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}\) is equal to

• A. \(1\)
• B. \(u\)
• C. \(-u\)
• D. \(0\)

Hint:

Remember Euler's theorem for homogeneous functions: \[ \boxed{ x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = nu } \] where \(n\) is the degree of homogeneity. If the degree is \(1\), the result is simply \[ \boxed{ x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = u. } \]

Answer 1

The Correct Option is B

Concept: The given function \[ u=\frac{xy}{x+y} \] is a homogeneous function. A function \(f(x,y)\) is said to be homogeneous of degree \(n\) if \[ f(\lambda x,\lambda y)=\lambda^n f(x,y). \] For homogeneous functions, Euler's theorem states \[ \boxed{ x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = nu } \] where \(n\) is the degree of homogeneity.

Step 1: Determine the degree of homogeneity.
Replace \(x\) by \(\lambda x\) and \(y\) by \(\lambda y\). \[ u(\lambda x,\lambda y) = \frac{(\lambda x)(\lambda y)} {\lambda x+\lambda y} \] \[ = \frac{\lambda^2xy} {\lambda(x+y)} \] \[ = \lambda\frac{xy}{x+y} \] \[ = \lambda u \] Hence, \[ \boxed{n=1} \]

Step 2: Apply Euler's theorem.
Since the degree of homogeneity is \(1\), \[ x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = 1\cdot u \] Therefore, \[ \boxed{ x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = u } \]

Step 3: Choose the correct option.
Thus, \[ \boxed{Option (B) is correct \]