Differentiate using first principle y=sqrt x

To differentiate \( y = \sqrt{x} \) using first principles, we use the definition of the derivative:

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

Given \( f(x) = \sqrt{x} \), let's substitute this into the formula:

1. **Substitute \( f(x) = \sqrt{x} \):**

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

2. **Multiply the numerator and denominator by the conjugate of the numerator \( \sqrt{x+h} + \sqrt{x} \):**

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

3. **Simplify the numerator using the difference of squares:**

\[
(\sqrt{x+h})^2 - (\sqrt{x})^2 = (x+h) - x = h
\]

Now, the expression becomes:

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

4. **Cancel \( h \) in the numerator and denominator:**

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

5. **Take the limit as \( h \to 0 \):**

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

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

Thus, the derivative of \( y = \sqrt{x} \) using first principles is:

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

Would you like more details on this? 

Here are 5 related questions:

1. How would you differentiate \( y = \frac{1}{x} \) using first principles?
2. How can you apply first principles to differentiate \( y = x^n \)?
3. What are the advantages of using the conjugate when simplifying expressions with square roots?
4. How do you differentiate \( y = \ln(x) \) using first principles?
5. Can you explain the concept of limits in the definition of a derivative?

**Tip:** The first principles method gives you an intuitive understanding of the derivative as the slope of the tangent line.

Learn how to differentiate y = sqrt(x) using first principles. This detailed solution walks through the use of limits, the difference of squares, and step-by-step calculations. Perfect for students in Grades 11-12 or early college-level calculus.

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