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