Ther derivative of f(x) = sec(X) tan(x) is

Options

A. `sec^(3) x tan x`
B. `sec (2sec^(2) x - 1)`
C. `sec x(tan^(2) x - 1)`
D. sec x

Answer

To find the derivative of the function \( f(x) = \sec(x) \tan(x) \), we will use the product rule of differentiation. The product rule states that if you have a function that is the product of two functions, say \( u \) and \( v \), then the derivative \( f'(x) \) is given by:

\[ f'(x) = u'v + uv' \]

### Step-by-step Solution:

1. **Identify \( u \) and \( v \)**:
Let \( u = \sec(x) \) and \( v = \tan(x) \).

2. **Differentiate \( u \) and \( v \)**:
- The derivative of \( u \) (i.e., \( \sec(x) \)) is:
\[ u' = \sec(x) \tan(x) \]
- The derivative of \( v \) (i.e., \( \tan(x) \)) is:
\[ v' = \sec^2(x) \]

3. **Apply the Product Rule**:
Now, we can apply the product rule:
\[ f'(x) = u'v + uv' \]
Substituting the values we found:
\[ f'(x) = (\sec(x) \tan(x)) \tan(x) + \sec(x) (\sec^2(x)) \]

4. **Simplify the Expression**:
- The first term becomes:
\[ \sec(x) \tan^2(x) \]
- The second term becomes:
\[ \sec^3(x) \]
- Therefore, we can combine these:
\[ f'(x) = \sec(x) \tan^2(x) + \sec^3(x) \]

5. **Factor the Expression**:
We can factor out \( \sec(x) \):
\[ f'(x) = \sec(x) (\tan^2(x) + \sec^2(x)) \]

6. **Use the Identity**:
We know from trigonometric identities that:
\[ \tan^2(x) + 1 = \sec^2(x) \]
Therefore, we can express \( \tan^2(x) \) as:
\[ \tan^2(x) = \sec^2(x) - 1 \]
Substituting this into our expression gives:
\[ f'(x) = \sec(x) ((\sec^2(x) - 1) + \sec^2(x)) \]
Simplifying this yields:
\[ f'(x) = \sec(x) (2\sec^2(x) - 1) \]

### Final Result:
Thus, the derivative of \( f(x) = \sec(x) \tan(x) \) is:
\[ f'(x) = \sec(x) (2\sec^2(x) - 1) \]