# Given x = cost y=sin2t, how do you find the dy/dx terms parameter t and find the values parameter t points dy/dx = 0?

graph{2xsqrt(1-x^2) [-2.527, 2.473, -1.11, 1.39]}

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To find ( \frac{dy}{dx} ) in terms of the parameter ( t ), where ( x = \cos(t) ) and ( y = \sin^2(t) ), you need to use the chain rule. First, express ( y ) explicitly as a function of ( x ) by substituting ( x = \cos(t) ) into the equation for ( y ). Then, differentiate ( y ) with respect to ( x ) using the chain rule. Once you have ( \frac{dy}{dx} ) in terms of ( x ), rewrite ( x ) in terms of ( t ), and simplify the expression.

To find the values of ( t ) where ( \frac{dy}{dx} = 0 ), set the expression for ( \frac{dy}{dx} ) equal to zero and solve for ( t ). These values of ( t ) will be the parameter points where the slope of the tangent line to the curve is zero.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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