Differentiate from first principles:
5. `sqrt( x+1), x gt -1`

Answer

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

### Step 1: Write 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 \( f(x) \) and \( f(x + h) \)
Here, we have:
\[ f(x) = \sqrt{x + 1} \]
Now, we need to find \( f(x + h) \):
\[ f(x + h) = \sqrt{(x + h) + 1} = \sqrt{x + h + 1} \]
Substituting these into the derivative formula gives:
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x + h + 1} - \sqrt{x + 1}}{h} \]

### Step 3: Rationalize the numerator
To simplify the expression, we will rationalize the numerator:
\[ f'(x) = \lim_{h \to 0} \frac{\left(\sqrt{x + h + 1} - \sqrt{x + 1}\right) \cdot \left(\sqrt{x + h + 1} + \sqrt{x + 1}\right)}{h \cdot \left(\sqrt{x + h + 1} + \sqrt{x + 1}\right)} \]
This gives:
\[ = \lim_{h \to 0} \frac{(x + h + 1) - (x + 1)}{h \cdot \left(\sqrt{x + h + 1} + \sqrt{x + 1}\right)} \]

### Step 4: Simplify the expression
The numerator simplifies to:
\[ = \lim_{h \to 0} \frac{h}{h \cdot \left(\sqrt{x + h + 1} + \sqrt{x + 1}\right)} \]
Now, we can cancel \( h \) in the numerator and denominator (as long as \( h \neq 0 \)):
\[ = \lim_{h \to 0} \frac{1}{\sqrt{x + h + 1} + \sqrt{x + 1}} \]

### Step 5: Evaluate the limit
Now we can substitute \( h = 0 \):
\[ = \frac{1}{\sqrt{x + 1} + \sqrt{x + 1}} = \frac{1}{2\sqrt{x + 1}} \]

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

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