# What is the slope of the polar curve #f(theta) = theta - sec^2theta+costhetasin^3theta # at #theta = (5pi)/6#?

From the reference Polar Curve Tangents :

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To find the slope of the polar curve ( f(\theta) = \theta - \sec^2(\theta) + \cos(\theta)\sin^3(\theta) ) at ( \theta = \frac{5\pi}{6} ), you need to find the first derivative of the function with respect to ( \theta ) and then evaluate it at ( \theta = \frac{5\pi}{6} ). The first derivative of ( f(\theta) ) with respect to ( \theta ) is ( f'(\theta) = 1 - \sec^2(\theta) - 2\sec(\theta)\tan(\theta) - \sin(\theta)\cos^3(\theta) + 3\sin^2(\theta)\cos^2(\theta) ). Evaluate ( f'(\theta) ) at ( \theta = \frac{5\pi}{6} ) to find the slope of the curve at that point.

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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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