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To solve this problem, we need to find the slope of the tangent line to the function \( t(x) = \csc(3x) + \cot(3x) \) at \( x = \frac{\pi}{4} \). The slope of the tangent line is given by the derivative of the function at that point.

### Step 1: Find the derivative of the function
The function is:
\[
t(x) = \csc(3x) + \cot(3x)
\]

To differentiate \( t(x) \), we use the chain rule along with the derivatives of the cosecant and cotangent functions:
- The derivative of \( \csc(u) \) is \( -\csc(u) \cot(u) \) where \( u = 3x \).
- The derivative of \( \cot(u) \) is \( -\csc^2(u) \) where \( u = 3x \).

So, applying the chain rule:

\[
t'(x) = \frac{d}{dx}[\csc(3x)] + \frac{d}{dx}[\cot(3x)]
\]
\[
t'(x) = -\csc(3x) \cdot \cot(3x) \cdot 3 - \csc^2(3x) \cdot 3
\]
\[
t'(x) = -3\csc(3x) \cot(3x) - 3\csc^2(3x)
\]

### Step 2: Evaluate the derivative at \( x = \frac{\pi}{4} \)
Now, we substitute \( x = \frac{\pi}{4} \) into the derivative:

1. Calculate \( 3x \) at \( x = \frac{\pi}{4} \):
   \[
   3 \cdot \frac{\pi}{4} = \frac{3\pi}{4}
   \]

2. Evaluate \( \csc\left(\frac{3\pi}{4}\right) \) and \( \cot\left(\frac{3\pi}{4}\right) \):
   - \( \sin\left(\frac{3\pi}{4}\right) = \sin\left(\pi - \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \), so \( \csc\left(\frac{3\pi}{4}\right) = \frac{1}{\sin\left(\frac{3\pi}{4}\right)} = \sqrt{2} \).
   - \( \cot\left(\frac{3\pi}{4}\right) = -\cot\left(\frac{\pi}{4}\right) = -1 \).

3. Substitute these values into the derivative:
   \[
   t'\left(\frac{\pi}{4}\right) = -3\left(\sqrt{2}\right)\left(-1\right) - 3\left(\sqrt{2}\right)^2
   \]
   Simplifying:
   \[
   t'\left(\frac{\pi}{4}\right) = 3\sqrt{2} - 6
   \]

Thus, the slope of the tangent line is \( 3\sqrt{2} - 6 \).

Would you like further details or clarifications? Let me know!

Here are 5 related questions to practice:
1. How do you find the derivative of a product of two trigonometric functions?
2. What is the derivative of \( \csc(x) \) in terms of sine and cosine functions?
3. Can you compute the slope of a tangent line for a different trigonometric function, like \( t(x) = \sec(x) + \tan(x) \)?
4. How would the chain rule apply if the argument inside \( \csc(3x) \) were more complex, like \( 5x + 2 \)?
5. What does the value of the slope tell us about the behavior of the function at \( x = \frac{\pi}{4} \)?

**Tip:** When differentiating composite trigonometric functions, always keep the chain rule in mind—especially when the argument involves a constant multiple of \( x \).

Find the slope of the tangent line to the graph of t(x) = csc(3x) + cot(3x), at the point where x = π/4.

Learn how to calculate the slope of the tangent line to the function t(x) = csc(3x) + cot(3x) at the point x = π/4. This detailed solution uses derivatives, the chain rule, and trigonometric identities to solve the problem. Suitable for high school calculus students.

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