The problem asks for the derivative of the function:

$f(x) = \sin^{-1}(14 \cos x)$

We need to differentiate this function with respect to $x$.

Step 1: Use the chain rule.

The function is composed of an outer function, the inverse sine $\sin^{-1}(u)$, and an inner function $u = 14 \cos(x)$. The chain rule tells us:

$f'(x) = \frac{d}{dx} \left[ \sin^{-1}(u) \right] \cdot \frac{du}{dx}$

The derivative of $\sin^{-1}(u)$ is:

$\frac{d}{dx} \left[ \sin^{-1}(u) \right] = \frac{1}{\sqrt{1 - u^2}}$

So:

$f'(x) = \frac{1}{\sqrt{1 - (14 \cos(x))^2}} \cdot \frac{d}{dx} \left[ 14 \cos(x) \right]$

Step 2: Differentiate the inner function.

The derivative of $14 \cos(x)$ with respect to $x$ is:

$\frac{d}{dx} \left[ 14 \cos(x) \right] = -14 \sin(x)$

Step 3: Combine the results.

Now substitute this into the derivative expression:

$f'(x) = \frac{-14 \sin(x)}{\sqrt{1 - (14 \cos(x))^2}}$

Step 4: Simplify the denominator.

We simplify the expression in the square root:

$(14 \cos(x))^2 = 196 \cos^2(x)$

Thus, the final expression for the derivative is:

$f'(x) = \frac{-14 \sin(x)}{\sqrt{1 - 196 \cos^2(x)}}$

This is the derivative of the function.

Would you like a more detailed explanation on any part? Here are some related questions:

1. How do you apply the chain rule to composite functions?
2. Why is the derivative of $\sin^{-1}(x)$ $\frac{1}{\sqrt{1 - x^2}}$?
3. Can you explain how to differentiate trigonometric functions like $\cos(x)$?
4. How does the domain of the inverse sine function affect the solution?
5. How can we check if the solution is correct?

Tip: When dealing with inverse trigonometric functions, it's helpful to remember the derivative formulas for each and how to apply the chain rule properly.