Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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**Problem:**

Given the function \( y = x \sqrt{x} \), find \( \frac{dy}{dx} \).

---

**Solution:**

To find the derivative of the function \( y = x \sqrt{x} \), we will apply the product rule and the chain rule of differentiation.

1. **Rewrite the function in exponent form:**

   \[
   y = x \cdot x^{1/2} = x^{1 + \frac{1}{2}} = x^{3/2}
   \]

2. **Differentiate using the power rule:**

   The power rule states that if \( y = x^n \), then \( \frac{dy}{dx} = n x^{n-1} \).

   \[
   y = x^{3/2}
   \]

   Using the power rule:

   \[
   \frac{dy}{dx} = \frac{3}{2} x^{\frac{3}{2} - 1} = \frac{3}{2} x^{\frac{1}{2}} = \frac{3}{2} \sqrt{x}
   \]

Thus, the derivative of \( y = x \sqrt{x} \) is:

\[
\frac{dy}{dx} = \frac{3}{2} \sqrt{x}
\]
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Transcribed Image Text:**Problem:** Given the function \( y = x \sqrt{x} \), find \( \frac{dy}{dx} \). --- **Solution:** To find the derivative of the function \( y = x \sqrt{x} \), we will apply the product rule and the chain rule of differentiation. 1. **Rewrite the function in exponent form:** \[ y = x \cdot x^{1/2} = x^{1 + \frac{1}{2}} = x^{3/2} \] 2. **Differentiate using the power rule:** The power rule states that if \( y = x^n \), then \( \frac{dy}{dx} = n x^{n-1} \). \[ y = x^{3/2} \] Using the power rule: \[ \frac{dy}{dx} = \frac{3}{2} x^{\frac{3}{2} - 1} = \frac{3}{2} x^{\frac{1}{2}} = \frac{3}{2} \sqrt{x} \] Thus, the derivative of \( y = x \sqrt{x} \) is: \[ \frac{dy}{dx} = \frac{3}{2} \sqrt{x} \]
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