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Let's break down the questions:

### 6a) 
We need to prove that if \( y = \csc(x) \), then:

\[
\frac{dy}{dx} = -\csc(x) \cot(x)
\]

**Solution:**

1. The derivative of \( y = \csc(x) \) can be found using the chain rule. We know:
   \[
   \csc(x) = \frac{1}{\sin(x)}
   \]
   
2. Now, differentiate \( \csc(x) \):
   \[
   \frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{\sin(x)} \right)
   \]

3. Using the quotient rule:
   \[
   \frac{dy}{dx} = \frac{0 \cdot \sin(x) - 1 \cdot \cos(x)}{\sin^2(x)} = -\frac{\cos(x)}{\sin^2(x)}
   \]

4. This simplifies to:
   \[
   \frac{dy}{dx} = -\csc(x) \cot(x)
   \]

Thus, the derivative is proven.

### 6b)
Given that \( x = \csc(6y) \), we need to find \( \frac{dy}{dx} \) in terms of \( x \).

**Solution:**

1. Start with \( x = \csc(6y) \), or equivalently:
   \[
   x = \frac{1}{\sin(6y)}
   \]

2. Differentiate both sides with respect to \( x \). Using implicit differentiation:
   \[
   \frac{d}{dx}\left(x\right) = \frac{d}{dx} \left( \frac{1}{\sin(6y)} \right)
   \]

3. The left side is simple:
   \[
   \frac{d}{dx} (x) = 1
   \]

4. For the right side, apply the chain rule:
   \[
   \frac{d}{dx} \left( \frac{1}{\sin(6y)} \right) = -\frac{\cos(6y)}{\sin^2(6y)} \cdot \frac{d}{dx}(6y) = -\csc(6y) \cot(6y) \cdot 6 \cdot \frac{dy}{dx}
   \]

5. Substitute \( \csc(6y) = x \) into the expression:
   \[
   1 = -6x \cot(6y) \frac{dy}{dx}
   \]

6. Now, solve for \( \frac{dy}{dx} \):
   \[
   \frac{dy}{dx} = \frac{-1}{6x \cot(6y)}
   \]

Thus, we have found the derivative in terms of \( x \).

---

Would you like more details on any of these steps?

### 5 Related Questions:
1. How is the derivative of \( \csc(x) \) related to the product rule?
2. Can you find the second derivative of \( y = \csc(x) \)?
3. How would you differentiate \( \csc(ax) \) for any constant \( a \)?
4. What is the significance of implicit differentiation in part (b)?
5. How do cotangent and cosecant derivatives relate to trigonometric identities?

**Tip**: For differentiating trigonometric functions, remembering key identities like \( \frac{d}{dx} \csc(x) = -\csc(x)\cot(x) \) can help speed up the process.

6a) Show that if y = cosec x then dy/dx = -cosec x cot x.
6b) Given x = cosec 6y, find dy/dx in terms of x.

Learn how to differentiate trigonometric functions like y = cosec(x) and x = cosec(6y) using the chain rule and implicit differentiation. This step-by-step guide walks you through key concepts such as the quotient rule and related trigonometric identities.

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