### Derivatives of Trigonometric Functions**Problem: Evaluate and simplify.**\[ \frac{d}{dx} \cos(9 \sec x) \]\[ \frac{d}{dx} \cos(9 \sec x) = \boxed{} \]### Solution Steps:To find the derivative of \(\cos(9 \sec x)\) with respect to \(x\), we will use the chain rule. Here's the detailed process:1. **Identify the outer function and the inner function:** In this case, the outer function is \(\cos\) and the inner function is \(9 \sec x\).2. **Find the derivative of the outer function:** The derivative of \(\cos(u)\) with respect to \(u\) is \(-\sin(u)\).3. **Find the derivative of the inner function:** The derivative of \(9 \sec x\) with respect to \(x\) is \(9 \sec x \tan x\).Using the chain rule, we multiply the derivative of the outer function by the derivative of the inner function:\[ \frac{d}{dx} \cos(9 \sec x) = -\sin(9 \sec x) \cdot \frac{d}{dx} (9 \sec x) \]\[ \frac{d}{dx} \cos(9 \sec x) = -\sin(9 \sec x) \cdot 9 \sec x \tan x \]\[ \frac{d}{dx} \cos(9 \sec x) = -9 \sin(9 \sec x) \sec x \tan x \]Hence, the simplified derivative of \(\cos(9 \sec x)\) with respect to \(x\) is:\[ \frac{d}{dx} \cos(9 \sec x) = -9 \sin(9 \sec x) \sec x \tan x \]### SummaryOn solving the given problem, the derivative evaluated at every step using the chain rule gives us \(-9 \sin(9 \sec x) \sec x \tan x\).

Name: ### Derivatives of Trigonometric Functions**Problem: Evaluate and sim
Uploaded: 2024-03-20T06:46:55.000Z
Channel: Bartleby
Description: ### Derivatives of Trigonometric Functions**Problem: Evaluate and simplify.**\[ \frac{d}{dx} \cos(9 \sec x) \]\[ \frac{d}{dx} \cos(9 \sec x) = \boxed{} \]### Solution Steps:To find the derivative of \(\cos(9 \sec x)\) with respect to \(x\), we will