write the value of `df ` if `f(x)=ln(x+1),x=1and delx=0.04`

Answer

To find the value of \( df \) for the function \( f(x) = \ln(x + 1) \) at \( x = 1 \) with \( \Delta x = 0.04 \), we will follow these steps:

### Step 1: Identify the function and its derivative
The function is given as:
\[ f(x) = \ln(x + 1) \]
We need to find the derivative \( f'(x) \).

### Step 2: Compute the derivative
To find \( f'(x) \), we use the derivative of the natural logarithm:
\[ f'(x) = \frac{d}{dx} \ln(x + 1) = \frac{1}{x + 1} \]

### Step 3: Evaluate the derivative at \( x = 1 \)
Now, we evaluate the derivative at \( x = 1 \):
\[ f'(1) = \frac{1}{1 + 1} = \frac{1}{2} \]

### Step 4: Use the formula for \( df \)
The differential \( df \) is given by the formula:
\[ df = f'(x) \cdot \Delta x \]
Substituting the values we have:
\[ df = f'(1) \cdot \Delta x = \frac{1}{2} \cdot 0.04 \]

### Step 5: Calculate \( df \)
Now, we perform the multiplication:
\[ df = \frac{1}{2} \cdot 0.04 = 0.02 \]

### Final Answer
Thus, the value of \( df \) is:
\[ \boxed{0.02} \]