Derivative of 1 /cosx

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 $\sec x \tan x$.