Use delta method to find the derivatives of the following : `cos(3x+5)`

Answer

To find the derivative of the function \( f(x) = \cos(3x + 5) \) using the delta method, we will follow these steps:

### Step 1: Define the function and the increment
Let \( f(x) = \cos(3x + 5) \). We need to find \( f(x + h) \):
\[ f(x + h) = \cos(3(x + h) + 5) = \cos(3x + 3h + 5) \]

### Step 2: Apply the definition of the derivative
The derivative \( f'(x) \) is defined as:
\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]
Substituting the values we found:
\[ f'(x) = \lim_{h \to 0} \frac{\cos(3x + 3h + 5) - \cos(3x + 5)}{h} \]

### Step 3: Use the cosine difference identity
We can use the identity \( \cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \):
Let \( A = 3x + 3h + 5 \) and \( B = 3x + 5 \):
\[ f'(x) = \lim_{h \to 0} \frac{-2 \sin\left(\frac{(3x + 3h + 5) + (3x + 5)}{2}\right) \sin\left(\frac{(3x + 3h + 5) - (3x + 5)}{2}\right)}{h} \]

### Step 4: Simplify the expression
Calculating the averages:
\[ \frac{(3x + 3h + 5) + (3x + 5)}{2} = \frac{6x + 3h + 10}{2} = 3x + \frac{3h}{2} + 5 \]
\[ \frac{(3x + 3h + 5) - (3x + 5)}{2} = \frac{3h}{2} \]
Thus, we can rewrite:
\[ f'(x) = \lim_{h \to 0} \frac{-2 \sin\left(3x + \frac{3h}{2} + 5\right) \sin\left(\frac{3h}{2}\right)}{h} \]

### Step 5: Factor out the limit
We can separate the limit:
\[ f'(x) = -2 \lim_{h \to 0} \sin\left(3x + \frac{3h}{2} + 5\right) \cdot \lim_{h \to 0} \frac{\sin\left(\frac{3h}{2}\right)}{h} \]
Using the standard limit \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \):
\[ \lim_{h \to 0} \frac{\sin\left(\frac{3h}{2}\right)}{h} = \lim_{h \to 0} \frac{\sin\left(\frac{3h}{2}\right)}{\frac{3h}{2}} \cdot \frac{3}{2} = \frac{3}{2} \]

### Step 6: Substitute back into the derivative
Now substituting back:
\[ f'(x) = -2 \sin(3x + 5) \cdot \frac{3}{2} \]
Thus, we get:
\[ f'(x) = -3 \sin(3x + 5) \]

### Final Result
The derivative of \( f(x) = \cos(3x + 5) \) is:
\[ f'(x) = -3 \sin(3x + 5) \]