Answer:
[tex]\displaystyle \frac{sin(\sqrt{x})}{4x}[/tex]
General Formulas and Concepts:
Algebra I
- Exponential Rule [Rewrite]: [tex]\displaystyle b^{-m} = \frac{1}{b^m}[/tex]
- Exponential Rule [Root Rewrite]: [tex]\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}[/tex]
Calculus
Derivatives
Derivative Notation
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Chain Rule]: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
Integrals
Integration Rule [Fundamental Theorem of Calculus 2]: [tex]\displaystyle \frac{d}{dx}[\int\limits^x_a {f(t)} \, dt] = f(x)[/tex]
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle g(t) = \int\limits^{\sqrt{x}}_1 {\frac{sin(t)}{2t}} \, dt[/tex]
Step 2: Differentiate
- Fundamental Theorem of Calculus 2 [Derivative Rule - Chain Rule]: [tex]\displaystyle g'(x) = \frac{sin(\sqrt{x})}{2\sqrt{x}} \cdot \frac{d}{dx}[\sqrt{x}][/tex]
- Rewrite [Exponential Rule - Root Rewrite]: [tex]\displaystyle g'(x) = \frac{sin(\sqrt{x})}{2\sqrt{x}} \cdot \frac{d}{dx} \bigg[ x^\bigg{\frac{1}{2}} \bigg][/tex]
- Basic Power Rule: [tex]\displaystyle g'(x) = \frac{sin(\sqrt{x})}{2\sqrt{x}} \cdot \frac{1}{2}x^\bigg{\frac{1}{2} - 1}[/tex]
- Simplify [Exponential Rule - Rewrite]: [tex]\displaystyle g'(x) = \frac{sin(\sqrt{x})}{2\sqrt{x}} \cdot \frac{1}{2x^\bigg{\frac{1}{2}}}[/tex]
- Rewrite [Exponential Rule - Root Rewrite]: [tex]\displaystyle g'(x) = \frac{sin(\sqrt{x})}{2\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}[/tex]
- Multiply: [tex]\displaystyle g'(x) = \frac{sin(\sqrt{x})}{4x}[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Book: College Calculus 10e
