Answer is: option2
\(-\frac{1}{\sqrt{3}}\)Solution:
Equation of the circle: \[ x^2 + y^2 = r^2 \]
Given:
\[ x^2 + y^2 = 4 \]Substituting \( x = 1 \):
\[ 1^2 + y^2 = 4 \] \[ y^2 = 3 \] \[ y = \pm \sqrt{3} \]Since \( f(1) = -\sqrt{3} \), we use implicit differentiation:
\[ 2x + 2y \frac{dy}{dx} = 0 \]Solving for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{x}{y} \]Substituting \( x = 1 \) and \( y = -\sqrt{3} \):
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{3}} \] Final Answer: \[ f'(1) = \boxed{-\frac{1}{\sqrt{3}}} \]