$$ x=\text{sec y },\frac{\differentialD y}{\differentialD x}= $$

Given the equation $ x = \sec(y) $, we want to find $ \frac{dy}{dx} $.

### Step-by-step solution:

1. Differentiate both sides of the equation $ x = \sec(y) $ with respect to $ x $:
   \[
   \frac{d}{dx} \left( x \right) = \frac{d}{dx} \left( \sec(y) \right)
   \]
   
2. The derivative of $ x $ with respect to $ x $ is simply 1:
   \[
   1 = \frac{d}{dx} \left( \sec(y) \right)
   \]

3. Use the chain rule to differentiate $ \sec(y) $ with respect to $ x $:
   \[
   \frac{d}{dx} \left( \sec(y) \right) = \sec(y) \tan(y) \cdot \frac{dy}{dx}
   \]
   
4. Now, set the derivatives equal to each other:
   \[
   1 = \sec(y) \tan(y) \cdot \frac{dy}{dx}
   \]

5. Solve for $ \frac{dy}{dx} $:
   \[
   \frac{dy}{dx} = \frac{1}{\sec(y) \tan(y)}
   \]

Since $ \sec(y) = x $, we substitute $ \sec(y) $ and $ \tan(y) $ in terms of $ x $:

6. Using the identity $ \tan^2(y) = \sec^2(y) - 1 $, we know that:
   \[
   \tan(y) = \sqrt{x^2 - 1}
   \]

7. Now substitute these into the expression for $ \frac{dy}{dx} $:
   \[
   \frac{dy}{dx} = \frac{1}{x \cdot \sqrt{x^2 - 1}}
   \]

Thus, the derivative is:
\[
\frac{dy}{dx} = \frac{1}{x \sqrt{x^2 - 1}}
\]

Would you like more details on the steps or any clarifications?

Here are 5 related questions to expand on this concept:
1. How would you differentiate $ y = \cos^{-1}(x) $ with respect to $ x $?
2. What is the derivative of $ y = \sec^{-1}(x) $?
3. How can you apply the chain rule to a function like $ z = \tan(\sec(y)) $?
4. What is the second derivative of $ y = \sec^{-1}(x) $?
5. How does the identity $ \sec^2(y) = 1 + \tan^2(y) $ help in simplifying derivatives?

**Tip:** Always consider trigonometric identities when simplifying derivatives involving trigonometric functions!

This solution provides a detailed step-by-step differentiation of x = sec(y) with respect to x using the chain rule and trigonometric identities. Explore how to simplify derivatives involving trigonometric functions and apply identities like sec^2(y) = 1 + tan^2(y). Suitable for advanced high school students and early college calculus learners.

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