Evaluate

𝑑𝑦

𝑑𝑥

(𝑥 + 1)(𝑥 + √𝑥) at x = 9

We need to find the derivative of the product \( (x + 1)(x + \sqrt{x}) \) and evaluate it at \( x = 9 \). This requires the use of the **product rule** for differentiation.

### Step 1: Apply the Product Rule
The product rule states:

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

Here, let:

- \( f(x) = (x + 1) \)
- \( g(x) = (x + \sqrt{x}) \)

### Step 2: Find the derivatives

1. **Derivative of \( f(x) = x + 1 \):**

\[
f'(x) = \frac{d}{dx}(x + 1) = 1
\]

2. **Derivative of \( g(x) = x + \sqrt{x} \):**

The derivative of \( x \) is \( 1 \), and the derivative of \( \sqrt{x} = x^{1/2} \) is \( \frac{1}{2} x^{-1/2} \).

So,

\[
g'(x) = \frac{d}{dx}(x + \sqrt{x}) = 1 + \frac{1}{2}x^{-1/2}
\]

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

### Step 3: Use the Product Rule

Now, apply the product rule:

\[
\frac{d}{dx} \left[ (x + 1)(x + \sqrt{x}) \right] = f'(x) \cdot g(x) + f(x) \cdot g'(x)
\]

Substitute the values of \( f(x), f'(x), g(x), \) and \( g'(x) \):

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

### Step 4: Simplify the expression

Simplify the right-hand side:

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

Distribute the terms:

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

Simplifying further:

\[
= 2x + \sqrt{x} + 1 + \frac{x+1}{2\sqrt{x}}
\]

### Step 5: Evaluate at \( x = 9 \)

Now, substitute \( x = 9 \) into the expression:

1. \( \sqrt{9} = 3 \)
2. \( \frac{x+1}{2\sqrt{x}} = \frac{9+1}{2\cdot 3} = \frac{10}{6} = \frac{5}{3} \)

So, the expression becomes:

\[
2(9) + 3 + 1 + \frac{5}{3}
\]

Simplify:

\[
= 18 + 3 + 1 + \frac{5}{3}
\]

Convert the whole numbers to fractions with a denominator of 3:

\[
= \frac{18 \cdot 3}{3} + \frac{3 \cdot 3}{3} + \frac{1 \cdot 3}{3} + \frac{5}{3}
\]
\[
= \frac{54}{3} + \frac{9}{3} + \frac{3}{3} + \frac{5}{3}
\]
\[
= \frac{71}{3}
\]

### Conclusion:
The derivative \( \frac{dy}{dx} \) of \( (x + 1)(x + \sqrt{x}) \) at \( x = 9 \) is \( \frac{71}{3} \), or approximately \( 23.67 \).

---

Would you like more explanation on any part of the process?

### Related Questions:
1. What is the product rule and how can it be applied to more complex functions?
2. How do you differentiate products of functions involving radicals?
3. How can you apply the chain rule in cases involving \( \sqrt{x} \)?
4. What other methods exist for finding derivatives when multiple terms are involved?
5. How do you evaluate derivatives for higher-order terms?

### Tip:
When differentiating products of functions, always apply the product rule carefully and simplify the expressions step by step to avoid errors!

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

Learn how to find the derivative of the function (x + 1)(x + √x) using the product rule and evaluate it at x = 9. This detailed solution covers differentiation concepts, including the product rule and the derivative of a square root function. Suitable for students in Grades 11-12.

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