Evaluate 𝑑𝑦𝑑𝑥(𝑥+1)(𝑥+ √𝑥) at x = 9

To evaluate \(\frac{dy}{dx}\) of the function \((x+1)(x + \sqrt{x})\) at \(x = 9\), we need to apply the product rule and then substitute \(x = 9\) into the derivative.

### Step-by-Step Solution:

Let:
\[
f(x) = (x+1)(x + \sqrt{x})
\]

1. **Differentiate using the product rule:**

   The product rule states that:
   \[
   \frac{d}{dx}[u(x) \cdot v(x)] = u'(x)v(x) + u(x)v'(x)
   \]
   where \( u(x) = (x+1) \) and \( v(x) = (x + \sqrt{x}) \).

2. **Differentiate each function:**

   - For \( u(x) = (x + 1) \):
     \[
     u'(x) = \frac{d}{dx}(x+1) = 1
     \]

   - For \( v(x) = (x + \sqrt{x}) \):
     \[
     v'(x) = \frac{d}{dx}(x + \sqrt{x}) = 1 + \frac{1}{2\sqrt{x}}
     \]

3. **Apply the product rule:**
   \[
   \frac{d}{dx}[(x+1)(x + \sqrt{x})] = u'(x)v(x) + u(x)v'(x)
   \]
   Substituting the expressions for \(u'(x)\), \(u(x)\), \(v(x)\), and \(v'(x)\):
   \[
   = 1 \cdot (x + \sqrt{x}) + (x+1) \cdot \left(1 + \frac{1}{2\sqrt{x}}\right)
   \]
   Simplifying:
   \[
   = (x + \sqrt{x}) + (x+1)\left(1 + \frac{1}{2\sqrt{x}}\right)
   \]
   \[
   = (x + \sqrt{x}) + (x+1) + \frac{x+1}{2\sqrt{x}}
   \]

4. **Substitute \(x = 9\):**

   - \( \sqrt{9} = 3 \)
   - \( \frac{1}{2\sqrt{9}} = \frac{1}{6} \)

   Now, substituting \(x = 9\) into the derivative expression:
   \[
   = (9 + 3) + (9 + 1) + \frac{9 + 1}{2 \times 3}
   \]
   \[
   = 12 + 10 + \frac{10}{6}
   \]
   Simplifying:
   \[
   = 12 + 10 + \frac{5}{3}
   \]
   Converting to a common denominator:
   \[
   = 22 + \frac{5}{3} = \frac{66}{3} + \frac{5}{3} = \frac{71}{3}
   \]

Thus, the derivative at \(x = 9\) is:
\[
\frac{71}{3} \approx 23.67
\]

Let me know if you'd like more details on any step!

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### Related Questions:
1. How is the product rule applied in more complex functions?
2. What is the geometric meaning of finding the derivative at a point?
3. How do you differentiate functions involving square roots?
4. What are some common mistakes when using the product rule?
5. Can you explain how to simplify fractions in the context of derivatives?

### Tip:
Always double-check if you're applying the product or chain rule when differentiating. Some functions require both!

Evaluate dy/dx of (x + 1)(x + √x) at x = 9

Learn how to evaluate the derivative of the function (x + 1)(x + √x) at x = 9 using the product rule. This problem covers key calculus concepts, including the differentiation of functions with square roots. Suitable for Grades 11-12 students.

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