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!