If `y=log((1)/(1+x))`,then `1+(dy)/(dx)=.......`

Options

A. `xe^(-y)`
B. `xe^(y)`
C. x
D. `-x`

Explanation

To solve the problem \( y = \log\left(\frac{1}{1+x}\right) \) and find \( 1 + \frac{dy}{dx} \), we will follow these steps:

### Step 1: Differentiate \( y \)
Given:
\[ y = \log\left(\frac{1}{1+x}\right) \]
We can rewrite this using the properties of logarithms:
\[ y = \log(1) - \log(1+x) = 0 - \log(1+x) = -\log(1+x) \]

Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = -\frac{d}{dx}(\log(1+x)) \]

Using the derivative of the logarithm:
\[ \frac{d}{dx}(\log(1+x)) = \frac{1}{1+x} \]
Thus,
\[ \frac{dy}{dx} = -\frac{1}{1+x} \]

### Step 2: Calculate \( 1 + \frac{dy}{dx} \)
Now we need to find:
\[ 1 + \frac{dy}{dx} = 1 - \frac{1}{1+x} \]

### Step 3: Simplify the expression
To simplify:
\[ 1 - \frac{1}{1+x} = \frac{(1+x) - 1}{1+x} = \frac{x}{1+x} \]

### Final Answer
Thus, the final result is:
\[ 1 + \frac{dy}{dx} = \frac{x}{1+x} \]