# How do you differentiate #r=5theta^2sectheta#?

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To differentiate the given polar equation ( r = 5\theta^2\sec(\theta) ), follow these steps:

- Recognize that ( r ) is a function of ( \theta ), indicating a polar equation.
- Apply the product rule for differentiating ( r = 5\theta^2\sec(\theta) ).
- Differentiate ( \theta^2 ) and ( \sec(\theta) ) separately.
- Substitute the polar coordinates ( r = \theta^2 ) and ( \theta = \theta ) into the formula for the product rule.
- Simplify the expression to get the final result.

Differentiating ( \theta^2 ): [ \frac{d}{d\theta}(\theta^2) = 2\theta ]

Differentiating ( \sec(\theta) ): [ \frac{d}{d\theta}(\sec(\theta)) = \sec(\theta)\tan(\theta) ]

Applying the product rule: [ \frac{d}{d\theta}(r) = \frac{d}{d\theta}(5\theta^2\sec(\theta)) ] [ \frac{dr}{d\theta} = 5(2\theta \sec(\theta) + \theta^2 \sec(\theta)\tan(\theta)) ]

Therefore, the derivative of ( r ) with respect to ( \theta ) is: [ \frac{dr}{d\theta} = 10\theta\sec(\theta) + 5\theta^2\sec(\theta)\tan(\theta) ]

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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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