Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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**Problem: Find the Derivative** Given the function: \[ y(x) = \sqrt{x^2 + 12} \] Calculate \(\frac{dy}{dx}\). **Solution:** Find \(\frac{dy}{dx} =\) [Blank space for solution] --- This problem presents a function \(y(x)\) and asks for the derivative \(\frac{dy}{dx}\). The function involves a square root, suggesting the use of chain rule for differentiation. ### Explanation: 1. **Identify the Outer Function**: The square root, \(\sqrt{u}\) (where \(u = x^2 + 12\)). 2. **Differentiate the Outer Function**: The derivative of \(\sqrt{u}\) is \(\frac{1}{2\sqrt{u}}\). 3. **Identify the Inner Function**: \(u = x^2 + 12\). 4. **Differentiate the Inner Function**: The derivative of \(x^2 + 12\) with respect to \(x\) is \(2x\). 5. **Apply the Chain Rule**: Multiply the derivative of the outer function by the derivative of the inner function. The full differentiation process results in: \[ \frac{dy}{dx} = \frac{1}{2\sqrt{x^2 + 12}} \times 2x = \frac{x}{\sqrt{x^2 + 12}} \]
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Transcribed Image Text:**Problem: Find the Derivative** Given the function: \[ y(x) = \sqrt{x^2 + 12} \] Calculate \(\frac{dy}{dx}\). **Solution:** Find \(\frac{dy}{dx} =\) [Blank space for solution] --- This problem presents a function \(y(x)\) and asks for the derivative \(\frac{dy}{dx}\). The function involves a square root, suggesting the use of chain rule for differentiation. ### Explanation: 1. **Identify the Outer Function**: The square root, \(\sqrt{u}\) (where \(u = x^2 + 12\)). 2. **Differentiate the Outer Function**: The derivative of \(\sqrt{u}\) is \(\frac{1}{2\sqrt{u}}\). 3. **Identify the Inner Function**: \(u = x^2 + 12\). 4. **Differentiate the Inner Function**: The derivative of \(x^2 + 12\) with respect to \(x\) is \(2x\). 5. **Apply the Chain Rule**: Multiply the derivative of the outer function by the derivative of the inner function. The full differentiation process results in: \[ \frac{dy}{dx} = \frac{1}{2\sqrt{x^2 + 12}} \times 2x = \frac{x}{\sqrt{x^2 + 12}} \]
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