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

Options

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

Answer

To find \( f'(x) \) given \( f''(x) = \sec^4 x + 4 \) with the initial 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: Break down 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:

\[ \int \sec^4 x \, dx = \tan x \sec x + C_1 \]

### Step 4: Integrate the constant

The integral of \( 4 \) is straightforward:

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

### Step 5: Combine the results

Combining the results from the integrals, we have:

\[ f'(x) = \tan x \sec x + 4x + C \]

where \( C = C_1 + C_2 \) is a constant of integration.

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

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

\[ f'(0) = \tan(0) \sec(0) + 4(0) + C = 0 \]

Since \( \tan(0) = 0 \) and \( \sec(0) = 1 \), we have:

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

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

Thus, the expression for \( f'(x) \) becomes:

\[ f'(x) = \tan x \sec x + 4x \]

### Summary of the solution

The final answer is:

\[ f'(x) = \tan x \sec x + 4x \]