If `f(x)=10cosx+(13+2x)sinx` then `f''(x)+f(x)=`

Options

A. `cosx`
B. `4cosx`
C. `sinx`
D. `4sinx`

Answer

To solve the problem, we need to find \( f''(x) + f(x) \) for the function \( f(x) = 10 \cos x + (13 + 2x) \sin x \).

### Step 1: Find the first derivative \( f'(x) \)

Given:
\[ f(x) = 10 \cos x + (13 + 2x) \sin x \]

We will differentiate each term separately:
1. The derivative of \( 10 \cos x \) is \( -10 \sin x \).
2. For \( (13 + 2x) \sin x \), we will use the product rule:
- Let \( u = 13 + 2x \) and \( v = \sin x \).
- Then, \( u' = 2 \) and \( v' = \cos x \).
- By the product rule, \( (uv)' = u'v + uv' \):
\[ \frac{d}{dx}((13 + 2x) \sin x) = (2) \sin x + (13 + 2x) \cos x \]

Combining these results:
\[ f'(x) = -10 \sin x + (2 \sin x + (13 + 2x) \cos x) \]
\[ f'(x) = (-10 + 2) \sin x + (13 + 2x) \cos x \]
\[ f'(x) = -8 \sin x + 13 \cos x + 2x \cos x \]

### Step 2: Find the second derivative \( f''(x) \)

Now we differentiate \( f'(x) \):
\[ f'(x) = -8 \sin x + 13 \cos x + 2x \cos x \]

Differentiating each term:
1. The derivative of \( -8 \sin x \) is \( -8 \cos x \).
2. The derivative of \( 13 \cos x \) is \( -13 \sin x \).
3. For \( 2x \cos x \), we again use the product rule:
- Let \( u = 2x \) and \( v = \cos x \).
- Then, \( u' = 2 \) and \( v' = -\sin x \).
- By the product rule:
\[ \frac{d}{dx}(2x \cos x) = (2) \cos x + (2x)(-\sin x) = 2 \cos x - 2x \sin x \]

Combining these results:
\[ f''(x) = -8 \cos x - 13 \sin x + (2 \cos x - 2x \sin x) \]
\[ f''(x) = (-8 + 2) \cos x - 13 \sin x - 2x \sin x \]
\[ f''(x) = -6 \cos x - 13 \sin x - 2x \sin x \]

### Step 3: Calculate \( f''(x) + f(x) \)

Now we add \( f(x) \) to \( f''(x) \):
\[ f(x) = 10 \cos x + (13 + 2x) \sin x \]

Adding \( f''(x) \) and \( f(x) \):
\[ f''(x) + f(x) = (-6 \cos x - 13 \sin x - 2x \sin x) + (10 \cos x + (13 + 2x) \sin x) \]
Combining like terms:
\[ = (-6 + 10) \cos x + (-13 + 13) \sin x + (-2x + 2x) \sin x \]
\[ = 4 \cos x + 0 + 0 \]
\[ = 4 \cos x \]

### Final Answer
Thus, we have:
\[ f''(x) + f(x) = 4 \cos x \]