# What is the slope of the tangent line of #x^2 + y^2 = 1/2 # at #x=1#?

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Geometrically

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To find the slope of the tangent line at a specific point on a curve, you first need to find the derivative of the curve with respect to x, and then evaluate it at the given x-coordinate. The derivative of the curve (x^2 + y^2 = \frac{1}{2}) with respect to x is (y' = -\frac{x}{y}). Substituting (x = 1) into this derivative, we get (y' = -\frac{1}{y}). To find the value of y at (x = 1), we can use the equation of the curve: (x^2 + y^2 = \frac{1}{2}) and solve for y. Solving this equation yields two possible values for y: (y = \sqrt{\frac{1}{2} - 1}) and (y = -\sqrt{\frac{1}{2} - 1}). Substituting these values into (y'), we get two slopes: (m_1 = -\sqrt{2} ) and (m_2 = \sqrt{2} ). Therefore, the slopes of the tangent lines at (x = 1) are ( -\sqrt{2} ) and ( \sqrt{2} ).

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