differentiate  using first principle  equals squareroot of  2x plus 1

To differentiate \( f(x) = \sqrt{2x + 1} \) using the first principles of differentiation, we apply the definition of the derivative:

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

In this case, \( f(x) = \sqrt{2x + 1} \), so let's substitute into the formula.

### Step 1: Write the difference quotient

\[
f'(x) = \lim_{h \to 0} \frac{\sqrt{2(x+h) + 1} - \sqrt{2x + 1}}{h}
\]

This becomes:

\[
f'(x) = \lim_{h \to 0} \frac{\sqrt{2x + 2h + 1} - \sqrt{2x + 1}}{h}
\]

### Step 2: Rationalize the numerator

To simplify this expression, we rationalize the numerator by multiplying both the numerator and the denominator by the conjugate of the numerator:

\[
f'(x) = \lim_{h \to 0} \frac{\left( \sqrt{2x + 2h + 1} - \sqrt{2x + 1} \right) \cdot \left( \sqrt{2x + 2h + 1} + \sqrt{2x + 1} \right)}{h \cdot \left( \sqrt{2x + 2h + 1} + \sqrt{2x + 1} \right)}
\]

Simplifying the numerator:

\[
= \lim_{h \to 0} \frac{\left( (2x + 2h + 1) - (2x + 1) \right)}{h \cdot \left( \sqrt{2x + 2h + 1} + \sqrt{2x + 1} \right)}
\]

This simplifies further to:

\[
= \lim_{h \to 0} \frac{2h}{h \cdot \left( \sqrt{2x + 2h + 1} + \sqrt{2x + 1} \right)}
\]

### Step 3: Cancel out the \( h \)

\[
= \lim_{h \to 0} \frac{2}{\sqrt{2x + 2h + 1} + \sqrt{2x + 1}}
\]

### Step 4: Take the limit as \( h \to 0 \)

As \( h \to 0 \), \( \sqrt{2x + 2h + 1} \to \sqrt{2x + 1} \), so:

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

### Final Answer:

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

Would you like more details on any of the steps? Here are some related questions for further exploration:

1. How do you differentiate a function involving square roots?
2. What is the significance of rationalizing the numerator in first principles?
3. How do you find the derivative of a general function using first principles?
4. Can you use this method for differentiating higher powers?
5. How does this compare to using standard differentiation rules?

**Tip:** Rationalizing the numerator is a common trick to simplify expressions involving square roots when using the first principle method.

differentiate using first principle equals squareroot of 2x plus 1

Explore the step-by-step process of differentiating the function f(x) = sqrt(2x + 1) using the first principles of differentiation. This solution covers key concepts like rationalizing the numerator and applying the limit as h approaches 0, suitable for Grades 11-12 or college-level calculus students.

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