`(1)/(sqrt(x+1)+sqrtx)`

Answer

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

### Step 1: Rationalize the Denominator
We start by rationalizing the denominator. We multiply the numerator and denominator by the conjugate of the denominator:

\[ f(x) = \frac{1}{\sqrt{x+1} + \sqrt{x}} \cdot \frac{\sqrt{x+1} - \sqrt{x}}{\sqrt{x+1} - \sqrt{x}} = \frac{\sqrt{x+1} - \sqrt{x}}{(\sqrt{x+1})^2 - (\sqrt{x})^2} \]

### Step 2: Simplify the Denominator
Now, we simplify the denominator using the difference of squares:

\[ (\sqrt{x+1})^2 - (\sqrt{x})^2 = (x + 1) - x = 1 \]

Thus, we have:

\[ f(x) = \sqrt{x+1} - \sqrt{x} \]

### Step 3: Differentiate the Function
Now we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx}(\sqrt{x+1}) - \frac{d}{dx}(\sqrt{x}) \]

Using the derivative formula \( \frac{d}{dx}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} \):

\[ f'(x) = \frac{1}{2\sqrt{x+1}} \cdot 1 - \frac{1}{2\sqrt{x}} \cdot 1 \]

This simplifies to:

\[ f'(x) = \frac{1}{2\sqrt{x+1}} - \frac{1}{2\sqrt{x}} \]

### Step 4: Combine the Terms
To combine these fractions, we find a common denominator:

\[ f'(x) = \frac{\sqrt{x} - \sqrt{x+1}}{2\sqrt{x}\sqrt{x+1}} \]

### Final Answer
Thus, the differential coefficient of the function \( f(x) = \frac{1}{\sqrt{x+1} + \sqrt{x}} \) is:

\[ f'(x) = \frac{\sqrt{x} - \sqrt{x+1}}{2\sqrt{x}\sqrt{x+1}} \]

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