differentiate w.r.t. x `(1)/(sqrt(x+1)+sqrtx)`

Answer

To differentiate the function \( y = \frac{1}{\sqrt{x+1} + \sqrt{x}} \) with respect to \( x \), we can follow these steps:

### Step 1: Rewrite the Function
We start with the function:
\[ y = \frac{1}{\sqrt{x+1} + \sqrt{x}} \]

### Step 2: Rationalize the Denominator
To differentiate this function more easily, we can rationalize the denominator. We multiply the numerator and denominator by the conjugate of the denominator:
\[ y = \frac{1}{\sqrt{x+1} + \sqrt{x}} \cdot \frac{\sqrt{x+1} - \sqrt{x}}{\sqrt{x+1} - \sqrt{x}} = \frac{\sqrt{x+1} - \sqrt{x}}{(\sqrt{x+1} + \sqrt{x})(\sqrt{x+1} - \sqrt{x})} \]

### Step 3: Simplify the Denominator
Using the difference of squares in the denominator:
\[ (\sqrt{x+1})^2 - (\sqrt{x})^2 = (x + 1) - x = 1 \]
Thus, the function simplifies to:
\[ y = \sqrt{x+1} - \sqrt{x} \]

### Step 4: Differentiate the Function
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(\sqrt{x+1}) - \frac{d}{dx}(\sqrt{x}) \]
Using the derivative of \( \sqrt{x} \), which is \( \frac{1}{2\sqrt{x}} \):
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x+1}} \cdot \frac{d}{dx}(x+1) - \frac{1}{2\sqrt{x}} \cdot \frac{d}{dx}(x) \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x+1}} - \frac{1}{2\sqrt{x}} \]

### Step 5: Factor Out Common Terms
We can factor out \( \frac{1}{2} \):
\[ \frac{dy}{dx} = \frac{1}{2} \left( \frac{1}{\sqrt{x+1}} - \frac{1}{\sqrt{x}} \right) \]

### Final Result
Thus, the derivative of the function \( y = \frac{1}{\sqrt{x+1} + \sqrt{x}} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{1}{2} \left( \frac{1}{\sqrt{x+1}} - \frac{1}{\sqrt{x}} \right) \]

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