Consider the function f ''(x) = sec^4 (x) + 4 with f(0) = 0 and f'(0) = 0 `f'(x)` equal to ?

Options

A. `tan x - (tan^(3)x)/(3) + 4x`
B. `tan x + (tan^(3)x)/(3) + 4x`
C. `tan x - (sec^(3)x)/(3) + 4x`
D. `- tan x - (tan^(3)x)/(3) + 4x`

Explanation

To find \( f'(x) \) given \( f''(x) = \sec^4(x) + 4 \) with the conditions \( f(0) = 0 \) and \( f'(0) = 0 \), we will follow these steps:

### Step 1: Integrate \( f''(x) \)

We start by integrating \( f''(x) \) to find \( f'(x) \):

\[ f'(x) = \int (\sec^4(x) + 4) \, dx \]

### Step 2: Separate the integral

We can separate the integral into two parts:

\[ f'(x) = \int \sec^4(x) \, dx + \int 4 \, dx \]

### Step 3: Integrate \( \sec^4(x) \)

The integral of \( \sec^4(x) \) can be computed using the identity:

\[ \int \sec^4(x) \, dx = \tan(x) + \frac{1}{3} \tan^3(x) + C_1 \]

### Step 4: Integrate the constant

Now, we integrate the constant:

\[ \int 4 \, dx = 4x + C_2 \]

### Step 5: Combine the results

Combining both integrals, we have:

\[ f'(x) = \tan(x) + \frac{1}{3} \tan^3(x) + 4x + C \]

where \( C = C_1 + C_2 \).

### Step 6: Apply the initial condition \( f'(0) = 0 \)

Now we apply the condition \( f'(0) = 0 \):

\[ f'(0) = \tan(0) + \frac{1}{3} \tan^3(0) + 4(0) + C = 0 \]

Since \( \tan(0) = 0 \), we have:

\[ 0 + 0 + 0 + C = 0 \implies C = 0 \]

### Step 7: Final expression for \( f'(x) \)

Thus, we find:

\[ f'(x) = \tan(x) + \frac{1}{3} \tan^3(x) + 4x \]

### Summary

The final answer for \( f'(x) \) is:

\[ f'(x) = \tan(x) + \frac{1}{3} \tan^3(x) + 4x \]