To find the derivative of $\frac{1}{\cos x}$, we can use the chain rule or rewrite the function using a negative exponent. Method 1: Using the Chain Rule Let $f(x) = \frac{1}{\cos x} = (\cos x)^{-1}$. Using the chain rule, the derivative is: $f'(x) = -1 (\cos x)^{-2} \cdot (-\sin x)$ $f'(x) = \frac{\sin x}{(\cos x)^2}$ $f'(x) = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x}$ $f'(x) = \tan x \sec x$ Method 2: Rewriting the function We know that $\sec x = \frac{1}{\cos x}$. So, the function is $\sec x$. The derivative of $\sec x$ is $\sec x \tan x$. $$ \frac{d}{dx} \left( \frac{1}{\cos x} \right) = \sec x \tan x $$ The derivative of $\frac{1}{\cos x}$ is $\sec x \tan x$. $\boxed{\sec x \tan x}$