Use the definition of derivative to calculate the f’(x) for f(x) = sqrtx Use the results to find the equation of the tangent line to the curve of f(x)=sqrtx at x=4



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?

---

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.

Use the definition of derivative to calculate the f’(x) for f(x) = sqrtx. Use the results to find the equation of the tangent line to the curve of f(x)=sqrtx at x=4.

Learn how to calculate the derivative of f(x) = sqrt(x) using the limit definition of the derivative. Follow step-by-step instructions to find the equation of the tangent line at x = 4. This solution covers core calculus concepts such as limits, derivatives, and tangent lines. Suitable for students in Grades 11-12.

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