Answer:
[tex]\displaystyle y' = -2 \cos \theta \cot (\sin \theta) \csc^2 (\sin \theta)[/tex]
General Formulas and Concepts:
Calculus
Differentiation
- 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]
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle y = \cot^2 (\sin \theta)[/tex]
Step 2: Differentiate
- Basic Power Rule [Derivative Rule - Chain Rule]: [tex]\displaystyle y' = 2 \cot (\sin \theta) \Big( \cot (\sin \theta) \Big)'[/tex]
- Trigonometric Differentiation [Derivative Rule - Chain Rule]: [tex]\displaystyle y' = -2 \cot (\sin \theta) \csc^2 (\sin \theta) (\sin \theta)'[/tex]
- Trigonometric Differentiation: [tex]\displaystyle y' = -2 \cos \theta \cot (\sin \theta) \csc^2 (\sin \theta)[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
