Obtain the differential coeffcient of the following :
`(1)/(2x+1)`

Answer

To find the differential coefficient of the function \( \frac{1}{2x+1} \), we will follow these steps:

### Step 1: Rewrite the function in a more convenient form
Given:
\[ y = \frac{1}{2x+1} \]

Rewrite it as:
\[ y = (2x+1)^{-1} \]

### Step 2: Apply the differentiation rule
We need to find \( \frac{dy}{dx} \). Using the chain rule for differentiation, we get:
\[ \frac{dy}{dx} = \frac{d}{dx} \left( (2x+1)^{-1} \right) \]

### Step 3: Differentiate using the power rule and chain rule
The power rule states that \( \frac{d}{dx} (u^n) = n \cdot u^{n-1} \cdot \frac{du}{dx} \), where \( u = 2x + 1 \) and \( n = -1 \).

\[ \frac{dy}{dx} = -1 \cdot (2x+1)^{-2} \cdot \frac{d}{dx}(2x+1) \]

### Step 4: Differentiate the inner function
\[ \frac{d}{dx}(2x+1) = 2 \]

### Step 5: Combine the results
\[ \frac{dy}{dx} = -1 \cdot (2x+1)^{-2} \cdot 2 \]
\[ \frac{dy}{dx} = -2 \cdot (2x+1)^{-2} \]

### Step 6: Simplify the expression
\[ \frac{dy}{dx} = \frac{-2}{(2x+1)^2} \]

So, the differential coefficient of \( \frac{1}{2x+1} \) is:
\[ \frac{dy}{dx} = \frac{-2}{(2x+1)^2} \]