Question

If \( u = x^4 + y^4 + 3x^2 y^2 \), the value of \( \left( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} \right)^{-1} \) is:

• A. \(-\frac{1}{3}\)
• B. \( \frac{1}{3} \)
• C. 3
• D. \(-3\)

Hint:

For partial derivatives involving terms with powers of \( x \) and \( y \), calculate the derivative with respect to each variable separately and then substitute into the expression to evaluate.

Answer 1

The Correct Option is C

Step 1: Find \( \frac{\partial u}{\partial x} \).
The given expression is: \[ u = x^4 + y^4 + 3x^2 y^2. \] We need to find the partial derivative of \(u\) with respect to \(x\). Differentiating with respect to \(x\), treating \(y\) as constant: \[ \frac{\partial u}{\partial x} = 4x^3 + 6xy^2. \] Step 2: Find \( \frac{\partial u}{\partial y} \).
Next, we find the partial derivative of \(u\) with respect to \(y\). Differentiating with respect to \(y\), treating \(x\) as constant: \[ \frac{\partial u}{\partial y} = 4y^3 + 6x^2 y. \] Step 3: Calculate \( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} \).
We now substitute the expressions for \( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \) into the desired expression: \[ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = x(4x^3 + 6xy^2) + y(4y^3 + 6x^2 y). \] Simplify: \[ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 4x^4 + 6x^2 y^2 + 4y^4 + 6x^2 y^2. \] Combine like terms: \[ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 4x^4 + 4y^4 + 12x^2 y^2. \] Notice that this expression is identical to \(u\), since: \[ u = x^4 + y^4 + 3x^2 y^2. \] Thus: \[ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 4u. \] Step 4: Find the inverse.
Finally, we take the inverse of the expression: \[ \left( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} \right)^{-1} = \frac{1}{4u}. \] Thus, the value of the expression is \( \frac{1}{4u} \).