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IM supposed to differentiate this, how do I get this problem to -tan u = 0

To differentiate the given function and simplify it, let's proceed step by step.

The function provided is:
\[ y = \sqrt{\pi - \theta} \cdot \tan(\pi - \theta) \cdot \sin(\theta) \]

Let's start by differentiating each component of the product separately using the product rule and chain rule.

1. **Product Rule**:
The product rule states that if you have a function \( y = u \cdot v \cdot w \), then:
\[ \frac{dy}{d\theta} = u'vw + uv'w + uvw' \]

Let's denote:
\[ u = \sqrt{\pi - \theta} \]
\[ v = \tan(\pi - \theta) \]
\[ w = \sin(\theta) \]

2. **Differentiate \( u \)**:
\[ u = \sqrt{\pi - \theta} \]
\[ \frac{du}{d\theta} = \frac{1}{2\sqrt{\pi - \theta}} \cdot (-1) = -\frac{1}{2\sqrt{\pi - \theta}} \]

3. **Differentiate \( v \)**:
\[ v = \tan(\pi - \theta) \]
Using the chain rule:
\[ \frac{dv}{d\theta} = \sec^2(\pi - \theta) \cdot (-1) = -\sec^2(\pi - \theta) \]

4. **Differentiate \( w \)**:
\[ w = \sin(\theta) \]
\[ \frac{dw}{d\theta} = \cos(\theta) \]

Now, applying the product rule:
\[ \frac{dy}{d\theta} = \frac{du}{d\theta} \cdot v \cdot w + u \cdot \frac{dv}{d\theta} \cdot w + u \cdot v \cdot \frac{dw}{d\theta} \]

Substitute the derivatives:
\[ \frac{dy}{d\theta} = \left( -\frac{1}{2\sqrt{\pi - \theta}} \right) \cdot \tan(\pi - \theta) \cdot \sin(\theta) + \sqrt{\pi - \theta} \cdot \left( -\sec^2(\pi - \theta) \right) \cdot \sin(\theta) + \sqrt{\pi - \theta} \cdot \tan(\pi - \theta) \cdot \cos(\theta) \]

Simplify each term:
1. \[ -\frac{\tan(\pi - \theta) \cdot \sin(\theta)}{2\sqrt{\pi - \theta}} \]
2. \[ -\sqrt{\pi - \theta} \cdot \sec^2(\pi - \theta) \cdot \sin(\theta) \]
3. \[ \sqrt{\pi - \theta} \cdot \tan(\pi - \theta) \cdot \cos(\theta) \]

Now, combine these terms to get the final derivative:
\[ \frac{dy}{d\theta} = -\frac{\tan(\pi - \theta) \cdot \sin(\theta)}{2\sqrt{\pi - \theta}} - \sqrt{\pi - \theta} \cdot \sec^2(\pi - \theta) \cdot \sin(\theta) + \sqrt{\pi - \theta} \cdot \tan(\pi - \theta) \cdot \cos(\theta) \]

To reach the simplified condition, \( -\tan u = 0 \):
For \( -\tan u = 0 \), it implies that \( \tan u = 0 \).

Given \( u = \pi - \theta \):
\[ \tan(\pi - \theta) = 0 \]
\[ \pi - \theta = n\pi \quad \text{(where \( n \) is an integer)} \]
\[ \theta = \pi - n\pi = (1 - n)\pi \]

Therefore, \( \theta \) must be an integer multiple of \( \pi \).

Would you like any further details or explanations?

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### Relative Questions
1. What is the chain rule in calculus?
2. How do you differentiate a function using the product rule?
3. What are the common trigonometric derivatives?
4. How does the secant function relate to the tangent function in differentiation?
5. Can you provide another example of a function using the product rule?
6. How do you simplify trigonometric functions involving \(\pi\)?
7. What are some common mistakes when differentiating trigonometric functions?
8. How can you verify your differentiation result?

### Tip
Always verify your differentiated function by substituting back into the original equation to check for consistency.

Learn how to differentiate the function y = sqrt(pi - theta) * tan(pi - theta) * sin(theta) step-by-step. This problem involves using the product rule and chain rule for differentiation, focusing on trigonometric functions and their derivatives. Suitable for undergraduate students and those studying calculus.

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