Answer is: option4

Solution:

Problem: Find the value(s) of \( \frac{dy}{dx} \) for the equation \( x^2 y + y^2 = 5 \) at \( y = 1 \).

Step 1: Differentiate the equation implicitly

The given equation is:

\[ x^2 y + y^2 = 5 \] To find \( \frac{dy}{dx} \), differentiate both sides implicitly with respect to \( x \):

\[ \frac{d}{dx} \left( x^2 y \right) + \frac{d}{dx} \left( y^2 \right) = \frac{d}{dx}(5) \]

Using the product rule and chain rule:

\[ 2x y + x^2 \frac{dy}{dx} + 2y \frac{dy}{dx} = 0 \]

Step 2: Solve for \( \frac{dy}{dx} \)

Now, group the terms involving \( \frac{dy}{dx} \):

\[ x^2 \frac{dy}{dx} + 2y \frac{dy}{dx} = -2xy \] Factor out \( \frac{dy}{dx} \): \[ \frac{dy}{dx}(x^2 + 2y) = -2xy \] Solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{-2xy}{x^2 + 2y} \]

Step 3: Substitute \( y = 1 \) into the equation

We are given \( y = 1 \). Substituting into the equation:

\[ \frac{dy}{dx} = \frac{-2x(1)}{x^2 + 2(1)} = \frac{-2x}{x^2 + 2} \]

Step 4: Determine the value(s) of \( x \) when \( y = 1 \)

We now find the corresponding values of \( x \) by substituting \( y = 1 \) into the original equation:

\[ x^2(1) + 1^2 = 5 \] \[ x^2 + 1 = 5 \] \[ x^2 = 4 \] \[ x = \pm 2 \]

Step 5: Calculate \( \frac{dy}{dx} \) for both values of \( x \)

Now, substitute \( x = 2 \) and \( x = -2 \) into the expression for \( \frac{dy}{dx} \):

For \( x = 2 \):

\[ \frac{dy}{dx} = \frac{-2(2)}{2^2 + 2} = \frac{-4}{4 + 2} = \frac{-4}{6} = -\frac{2}{3} \]

For \( x = -2 \):

\[ \frac{dy}{dx} = \frac{-2(-2)}{-2^2 + 2} = \frac{4}{4 + 2} = \frac{4}{6} = \frac{2}{3} \]

Final Answer:

The values of \( \frac{dy}{dx} \) are \( \pm \frac{2}{3} \).

Thus, the correct answer is (D) \( \pm \frac{2}{3} \).