Differentiate sec x

Let $u = \cos x$. Then $\frac{du}{dx} = -\sin x$.
The derivative of $u^{-1}$ with respect to $u$ is $-1 \cdot u^{-2}$.
So, using the chain rule $\frac{d}{dx} (u^{-1}) = \frac{d}{du} (u^{-1}) \cdot \frac{du}{dx}$:

$$\frac{d}{dx} ((\cos x)^{-1}) = -1 \cdot (\cos x)^{-2} \cdot (-\sin x)$$

$$-1 \cdot (\cos x)^{-2} \cdot (-\sin x) = \frac{\sin x}{(\cos x)^2} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x}$$ $$= \tan x \cdot \sec x$$

The derivative of $\sec x$ is $\boxed{\sec x \tan x}$.