If x^2 + y^2 = 4, what is dy/dx at the point (√3, 1)?

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Multiple Choice

If x^2 + y^2 = 4, what is dy/dx at the point (√3, 1)?

Explanation:
When a relation is given implicitly like x^2 + y^2 = 4, differentiate with respect to x to relate dy/dx to x and y. Differentiating both sides: 2x + 2y dy/dx = 0. Solve for dy/dx: dy/dx = -x/y. This gives the slope of the tangent line at any point on the circle, as long as y ≠ 0. At the point (√3, 1), substitute to get dy/dx = -(√3)/1 = -√3.

When a relation is given implicitly like x^2 + y^2 = 4, differentiate with respect to x to relate dy/dx to x and y. Differentiating both sides: 2x + 2y dy/dx = 0. Solve for dy/dx: dy/dx = -x/y. This gives the slope of the tangent line at any point on the circle, as long as y ≠ 0. At the point (√3, 1), substitute to get dy/dx = -(√3)/1 = -√3.

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