Differentiate from first principles:
7. `(1)/( sqrt (x+1))`

Answer

To differentiate the function \( f(x) = \frac{1}{\sqrt{x+1}} \) from first principles, we will follow these steps:

### Step 1: Write down the definition of the derivative from first principles.
The derivative of a function \( f(x) \) from first principles is given by:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Substitute \( f(x) \) into the formula.
For our function, we have:
\[ f(x) = \frac{1}{\sqrt{x+1}} \]
Thus,
\[ f(x+h) = \frac{1}{\sqrt{x+h+1}} \]
Now, substituting into the derivative formula:
\[ f'(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x+h+1}} - \frac{1}{\sqrt{x+1}}}{h} \]

### Step 3: Simplify the expression.
To simplify the expression, we need a common denominator:
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x+1} - \sqrt{x+h+1}}{h \cdot \sqrt{x+h+1} \cdot \sqrt{x+1}} \]

### Step 4: Rationalize the numerator.
To eliminate the square roots in the numerator, we multiply by the conjugate:
\[ f'(x) = \lim_{h \to 0} \frac{(\sqrt{x+1} - \sqrt{x+h+1})(\sqrt{x+1} + \sqrt{x+h+1})}{h \cdot \sqrt{x+h+1} \cdot \sqrt{x+1} \cdot (\sqrt{x+1} + \sqrt{x+h+1})} \]
This simplifies to:
\[ f'(x) = \lim_{h \to 0} \frac{(x+1) - (x+h+1)}{h \cdot \sqrt{x+h+1} \cdot \sqrt{x+1} \cdot (\sqrt{x+1} + \sqrt{x+h+1})} \]
\[ = \lim_{h \to 0} \frac{-h}{h \cdot \sqrt{x+h+1} \cdot \sqrt{x+1} \cdot (\sqrt{x+1} + \sqrt{x+h+1})} \]

### Step 5: Cancel \( h \) and evaluate the limit.
Cancelling \( h \) from the numerator and denominator:
\[ f'(x) = \lim_{h \to 0} \frac{-1}{\sqrt{x+h+1} \cdot \sqrt{x+1} \cdot (\sqrt{x+1} + \sqrt{x+h+1})} \]
Now, as \( h \to 0 \), \( \sqrt{x+h+1} \to \sqrt{x+1} \):
\[ f'(x) = \frac{-1}{\sqrt{x+1} \cdot \sqrt{x+1} \cdot (2\sqrt{x+1})} \]
\[ = \frac{-1}{2(x+1)} \]

### Final Result:
Thus, the derivative of \( f(x) = \frac{1}{\sqrt{x+1}} \) is:
\[ f'(x) = -\frac{1}{2(x+1)^{3/2}} \]

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