The first principle of derivative states that the derivative of a function $f(x)$ at a point $x$ is given by:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

For the given function:

$f(x) = \sqrt{x+1}$

Step 1: Apply the First Principle Formula

$f'(x) = \lim_{h \to 0} \frac{\sqrt{(x+h)+1} - \sqrt{x+1}}{h}$

$= \lim_{h \to 0} \frac{\sqrt{x+h+1} - \sqrt{x+1}}{h}$

Step 2: Multiply by the Conjugate

To simplify the numerator, multiply and divide by the conjugate:

$\frac{\sqrt{x+h+1} - \sqrt{x+1}}{h} \times \frac{\sqrt{x+h+1} + \sqrt{x+1}}{\sqrt{x+h+1} + \sqrt{x+1}}$

$= \lim_{h \to 0} \frac{(x+h+1) - (x+1)}{h (\sqrt{x+h+1} + \sqrt{x+1})}$

$= \lim_{h \to 0} \frac{x+h+1 - x - 1}{h (\sqrt{x+h+1} + \sqrt{x+1})}$

$= \lim_{h \to 0} \frac{h}{h (\sqrt{x+h+1} + \sqrt{x+1})}$

Cancel $h$ in numerator and denominator:

$= \lim_{h \to 0} \frac{1}{\sqrt{x+h+1} + \sqrt{x+1}}$

Step 3: Evaluate the Limit

As $h \to 0$, $\sqrt{x+h+1} \to \sqrt{x+1}$, so:

$f'(x) = \frac{1}{2\sqrt{x+1}}$

Final Answer:

$f'(x) = \frac{1}{2\sqrt{x+1}}$

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Here are 5 related questions:

1. Find the derivative of $f(x) = \sqrt{x}$ using the first principle.
2. Derive $f(x) = \frac{1}{\sqrt{x+1}}$ using the first principle.
3. Compute the derivative of $f(x) = (x+1)^{1/2}$ using the power rule.
4. Prove that the derivative of $f(x) = x^n$ is $nx^{n-1}$ using first principles.
5. Find the second derivative of $f(x) = \sqrt{x+1}$.

Tip: When simplifying limits, using the conjugate is a powerful technique to handle square roots!