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

We are tasked with finding the derivative of the function \( y = (x + 1)(x + \sqrt{x}) \) with respect to \( x \), and then evaluating it at \( x = 9 \).

### Step 1: Use the Product Rule

The product rule states that if \( y = f(x)g(x) \), then

\[
\frac{dy}{dx} = f'(x)g(x) + f(x)g'(x)
\]

Here, we can set:
- \( f(x) = (x + 1) \)
- \( g(x) = (x + \sqrt{x}) \)

### Step 2: Find the Derivatives of \( f(x) \) and \( g(x) \)

1. \( f(x) = x + 1 \), so the derivative of \( f(x) \) is:
   \[
   f'(x) = 1
   \]

2. \( g(x) = x + \sqrt{x} \), so the derivative of \( g(x) \) is:
   \[
   g'(x) = 1 + \frac{1}{2\sqrt{x}}
   \]

### Step 3: Apply the Product Rule

Now, using the product rule:

\[
\frac{dy}{dx} = f'(x)g(x) + f(x)g'(x)
\]

Substitute \( f(x) \), \( g(x) \), \( f'(x) \), and \( g'(x) \):

\[
\frac{dy}{dx} = 1 \cdot (x + \sqrt{x}) + (x + 1) \cdot \left(1 + \frac{1}{2\sqrt{x}}\right)
\]

Simplifying:

\[
\frac{dy}{dx} = (x + \sqrt{x}) + (x + 1) \left(1 + \frac{1}{2\sqrt{x}}\right)
\]

Expand \( (x + 1)(1 + \frac{1}{2\sqrt{x}}) \):

\[
(x + 1) \cdot \left(1 + \frac{1}{2\sqrt{x}}\right) = (x + 1) + \frac{x + 1}{2\sqrt{x}}
\]

Thus, we have:

\[
\frac{dy}{dx} = x + \sqrt{x} + (x + 1) + \frac{x + 1}{2\sqrt{x}}
\]

### Step 4: Evaluate at \( x = 9 \)

Substitute \( x = 9 \) into the expression:

1. \( x + \sqrt{x} = 9 + 3 = 12 \)
2. \( x + 1 = 9 + 1 = 10 \)
3. \(

Evaluate 𝑑𝑦/𝑑𝑥 for the function (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 product rule and the derivative of a square root function, suitable for students in Grades 11-12 or introductory calculus courses.

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