Let 𝑓(𝑥) = √𝑥. Answer the following questions:
a) Use the limit definition of the derivative to find 𝑓
′(𝑥). b) Find the equation of the tangent line at x = 9

To solve this problem, we'll go step by step through both parts:

### Part (a) - Finding the derivative of \( f(x) = \sqrt{x} \) using the limit definition of the derivative

The limit definition of the derivative is given by:

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

For \( f(x) = \sqrt{x} \), we substitute this into the formula:

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

To simplify this expression, we multiply both the numerator and denominator by the conjugate of the numerator, which is \( \sqrt{x+h} + \sqrt{x} \):

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

This simplifies the numerator using the difference of squares:

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

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

The \( h \)'s cancel out:

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

Now, taking the limit as \( h \to 0 \), we get:

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

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

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

---

### Part (b) - Finding the equation of the tangent line at \( x = 9 \)

To find the equation of the tangent line, we need two things:
1. The slope of the tangent line at \( x = 9 \), which is given by \( f'(9) \).
2. The point on the curve at \( x = 9 \), which is \( f(9) \).

First, find the slope of the tangent line:
\[
f'(9) = \frac{1}{2\sqrt{9}} = \frac{1}{2 \times 3} = \frac{1}{6}
\]

Next, find the point on the curve at \( x = 9 \):
\[
f(9) = \sqrt{9} = 3
\]

So the point is \( (9, 3) \), and the slope of the tangent line is \( \frac{1}{6} \).

Now, using the point-slope form of the equation of a line:

\[
y - y_1 = m(x - x_1)
\]

where \( m = \frac{1}{6} \), \( x_1 = 9 \), and \( y_1 = 3 \). Substituting these values:

\[
y - 3 = \frac{1}{6}(x - 9)
\]

Simplifying:

\[
y - 3 = \frac{1}{6}x - \frac{9}{6}
\]

\[
y = \frac{1}{6}x - \frac{3}{2} + 3
\]

\[
y = \frac{1}{6}x + \frac{3}{2}
\]

Thus, the equation of the tangent line at \( x = 9 \) is:

\[
y = \frac{1}{6}x + \frac{3}{2}
\]

---

Would you like a deeper explanation for any of the steps?

Here are 5 related questions to consider:
1. What is the geometric interpretation of the derivative at \( x = 9 \)?
2. How would the derivative change if \( f(x) = x^{1/3} \)?
3. Can you use the power rule to differentiate \( f(x) = \sqrt{x} \) and compare it with the result from the limit definition?
4. What is the second derivative of \( f(x) = \sqrt{x} \), and what does it tell you about the concavity of the function?
5. How can you find the normal line at \( x = 9 \)?

**Tip**: When finding a derivative using the limit definition, multiplying by the conjugate often helps simplify square roots.

Let 𝑓(𝑥) = √𝑥. Answer the following questions: a) Use the limit definition of the derivative to find 𝑓′(𝑥). b) Find the equation of the tangent line at x = 9.

Learn how to find the derivative of f(x) = √x using the limit definition and how to determine the equation of the tangent line at x = 9. This step-by-step solution covers key calculus concepts, including the limit definition of the derivative and the point-slope form of a line.

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