To find the derivative of $f(x) = \sqrt{x}$ using the definition of a derivative (limits), we proceed as follows:

Definition of a derivative:

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

Step 1: Substitute $f(x) = \sqrt{x}$

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

Step 2: Rationalize the numerator

Multiply and divide by the conjugate of the numerator to simplify: $f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}$

This gives: $f'(x) = \lim_{h \to 0} \frac{(\sqrt{x+h})^2 - (\sqrt{x})^2}{h \cdot (\sqrt{x+h} + \sqrt{x})}$

Step 3: Simplify the numerator

$(\sqrt{x+h})^2 - (\sqrt{x})^2 = (x+h) - x = h$ So: $f'(x) = \lim_{h \to 0} \frac{h}{h \cdot (\sqrt{x+h} + \sqrt{x})}$

Step 4: Cancel $h$ in the numerator and denominator

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

Step 5: Evaluate the limit as $h \to 0$

As $h \to 0$, $\sqrt{x+h} \to \sqrt{x}$. Thus: $f'(x) = \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}$

Final Answer:

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

Would you like me to elaborate on any step? 😊

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5 Related Questions:

1. What is the derivative of $f(x) = x^n$ for any $n$ using limits?
2. Can you use the same method to differentiate $f(x) = 1/\sqrt{x}$?
3. How is the derivative definition linked to the tangent of a curve?
4. What are some common functions where rationalizing the numerator is necessary for differentiation?
5. Why does rationalizing the numerator help in simplifying expressions?

Tip:

When finding derivatives using limits, rationalizing the numerator is often essential for simplifying expressions involving square roots.