https://file.aiquickdraw.com/mathbot/file/17269833608714znqbmkm.jpg

The expression provided in the image is:

\[
\frac{d}{dx} \left[ (\sqrt{x})(x - 1) \right]
\]

To differentiate this, we'll apply the **product rule** of differentiation, which is:

\[
\frac{d}{dx}(uv) = u'v + uv'
\]

Where \( u = \sqrt{x} = x^{1/2} \) and \( v = (x - 1) \).

### Step-by-step solution:

1. **Differentiate \( u = \sqrt{x} \):**
   \[
   u = x^{1/2} \quad \Rightarrow \quad u' = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}
   \]

2. **Differentiate \( v = x - 1 \):**
   \[
   v' = \frac{d}{dx}(x - 1) = 1
   \]

3. **Apply the product rule:**
   \[
   \frac{d}{dx} \left( \sqrt{x}(x - 1) \right) = u'v + uv'
   \]
   Substituting the values:
   \[
   = \left( \frac{1}{2\sqrt{x}} \right)(x - 1) + (\sqrt{x})(1)
   \]

4. **Simplify the expression:**
   \[
   = \frac{x - 1}{2\sqrt{x}} + \sqrt{x}
   \]

5. **Combine the terms:**
   \[
   \sqrt{x} = \frac{2x}{2\sqrt{x}}
   \]
   Therefore:
   \[
   \frac{x - 1}{2\sqrt{x}} + \frac{2x}{2\sqrt{x}} = \frac{x - 1 + 2x}{2\sqrt{x}} = \frac{3x - 1}{2\sqrt{x}}
   \]

Thus, the derivative is:

\[
\frac{d}{dx} \left[ (\sqrt{x})(x - 1) \right] = \frac{3x - 1}{2\sqrt{x}}
\]

Would you like further details or have any questions?

Here are five related questions you could explore:
1. How is the product rule applied in other functions involving square roots?
2. What is the significance of combining terms with common denominators in calculus?
3. Can you differentiate similar expressions involving cube roots or other fractional exponents?
4. How do you interpret this derivative in the context of a graph or application?
5. What are other scenarios where the product rule is commonly used?

**Tip:** Always simplify expressions as much as possible after applying differentiation rules to make further analysis easier.

Differentiate (sqrt(x))(x - 1) with respect to x.

This calculus problem demonstrates how to differentiate the expression (sqrt(x))(x - 1) using the product rule. Key concepts include the differentiation of square root functions and applying the product rule. Suitable for high school calculus students.

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