To calculate the derivative $f'(x)$ of $f(x) = \sqrt{x}$ using the definition of the derivative, we'll follow these steps:

Step 1: Definition of Derivative

The definition of the derivative is given by:

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

Step 2: Apply the Definition to $f(x) = \sqrt{x}$

For $f(x) = \sqrt{x}$, substitute into the definition:

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

Step 3: Simplify the Expression

To simplify, multiply the numerator and denominator by the conjugate of the numerator, $\sqrt{x+h} + \sqrt{x}$:

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

Since $(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x}) = (x+h) - x = h$, the expression simplifies to:

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

Now, cancel out $h$ from the numerator and denominator:

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

Step 4: Take the Limit as $h \to 0$

As $h \to 0$, $\sqrt{x+h}$ approaches $\sqrt{x}$. So, the expression becomes:

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

Thus, the derivative of $f(x) = \sqrt{x}$ is:

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

Step 5: Find the Tangent Line at $x = 4$

The slope of the tangent line at $x = 4$ is $f'(4)$. Using the derivative formula:

$f'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{4}$

Now, find the point on the curve at $x = 4$. Since $f(4) = \sqrt{4} = 2$, the point is $(4, 2)$.

Step 6: Equation of the Tangent Line

The equation of a tangent line is given by:

$y - y_1 = m(x - x_1)$

where $m$ is the slope, and $(x_1, y_1)$ is the point of tangency. Substituting $m = \frac{1}{4}$ and $(x_1, y_1) = (4, 2)$:

$y - 2 = \frac{1}{4}(x - 4)$

Simplifying:

$y = \frac{1}{4}x + 1$

Final Result:

• The derivative is $f'(x) = \frac{1}{2\sqrt{x}}$.
• The equation of the tangent line at $x = 4$ is:

$y = \frac{1}{4}x + 1$

Would you like further details or have any questions?

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Here are 5 related questions:

1. How do you find the tangent line at a different point for the same function $f(x) = \sqrt{x}$?
2. Can the same method be used to find the derivative of other functions like $f(x) = x^n$?
3. How does the concept of limits help in finding derivatives in general?
4. What happens to the slope of the tangent line as $x$ increases for $f(x) = \sqrt{x}$?
5. How do the properties of radicals affect the behavior of their derivatives?

Tip: The conjugate method is useful when simplifying expressions involving square roots during limit processes.