Answer is: option2

Solution:

Problem: The equation of a circle is \( x^2 + y^2 = 4 \). Find \( \frac{dy}{dx} \) at \( x = 1 \).

Step 1: Find \( y \) when \( x = 1 \)

\[ x^2 + y^2 = 4 \] Substituting \( x = 1 \): \[ 1^2 + y^2 = 4 \] \[ 1 + y^2 = 4 \] \[ y^2 = 3 \] \[ y = \sqrt{3} \] Thus, when \( x = 1 \), \( y = \sqrt{3} \).

Step 2: Differentiate the equation \( x^2 + y^2 = 4 \) implicitly

Differentiating both sides: \[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(4) \] \[ 2x + 2y \frac{dy}{dx} = 0 \]

Step 3: Solve for \( \frac{dy}{dx} \)

Solving for \( \frac{dy}{dx} \): \[ 2y \frac{dy}{dx} = -2x \] \[ \frac{dy}{dx} = -\frac{x}{y} \]

Step 4: Substitute \( x = 1 \) and \( y = \sqrt{3} \)

Substituting the values into the derivative: \[ \frac{dy}{dx} = -\frac{1}{\sqrt{3}} \] Thus, the derivative \( \frac{dy}{dx} \) at \( x = 1 \) is \( -\frac{1}{\sqrt{3}} \).