Respuesta :
Answer:
[tex]\displaystyle f'(x) = \frac{-5sin(5x)}{2\sqrt{cos(5x)}}[/tex]
General Formulas and Concepts:
Algebra I
Functions
- Function Notation
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
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]: [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]
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]
Trig Derivatives
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle f(x) = \sqrt{cos(5x)}[/tex]
Step 2: Differentiate
- Rewrite [Exponential Rule - Root Rewrite]: [tex]\displaystyle f(x) = [cos(5x)]^\bigg{\frac{1}{2}}[/tex]
- Derivative Rule [Chain Rule]: [tex]\displaystyle f'(x) = \frac{d}{dx} \bigg[ [cos(5x)]^\bigg{\frac{1}{2}} \bigg] \cdot \frac{d}{dx}[cos(5x)] \cdot \frac{d}{dx}[5x][/tex]
- Rewrite [Derivative Property - Multiplied Constant]: [tex]\displaystyle f'(x) = \frac{d}{dx} \bigg[ [cos(5x)]^\bigg{\frac{1}{2}} \bigg] \cdot \frac{d}{dx}[cos(5x)] \cdot 5\frac{d}{dx}[x][/tex]
- Basic Power Rule: [tex]\displaystyle f'(x) = \frac{1}{2}[cos(5x)]^\bigg{\frac{1}{2} - 1} \cdot \frac{d}{dx}[cos(5x)] \cdot 5x^{1 - 1}[/tex]
- Simplify: [tex]\displaystyle f'(x) = \frac{5}{2}[cos(5x)]^\bigg{\frac{-1}{2}} \cdot \frac{d}{dx}[cos(5x)][/tex]
- Trig Derivative: [tex]\displaystyle f'(x) = \frac{5}{2}[cos(5x)]^\bigg{\frac{-1}{2}} \cdot -sin(5x)[/tex]
- Simplify: [tex]\displaystyle f'(x) = \frac{-5sin(5x)}{2}[cos(5x)]^\bigg{\frac{-1}{2}}[/tex]
- Rewrite [Exponential Rule - Rewrite]: [tex]\displaystyle f'(x) = \frac{-5sin(5x)}{2[cos(5x)]^\bigg{\frac{1}{2}}}[/tex]
- Rewrite [Exponential Rule - Root Rewrite]: [tex]\displaystyle f'(x) = \frac{-5sin(5x)}{2\sqrt{cos(5x)}}[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Derivatives
Book: College Calculus 10e
