Answer:
\(\displaystyle \frac{dy}{dx} = \frac{-\csc x(\cot x \sin x + \cos x)}{3\sin^2 x}\)
General Formulas and Concepts:
Calculus
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]: \(\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)\)
Derivative Rule [Quotient Rule]: \(\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}\)
Step-by-step explanation:
Step 1: Define
Identify
\(\displaystyle y = \frac{\csc x}{3\sin x}\)
Step 2: Differentiate
- Derivative Property [Multiplied Constant]: \(\displaystyle y' = \frac{1}{3} \frac{d}{dx}[\frac{\csc x}{\sin x}]\)
- Derivative Rule [Quotient Rule]: \(\displaystyle y' = \frac{1}{3} \bigg( \frac{(\csc x)' \sin x - \csc x (\sin x)'}{\sin^2 x} \bigg)\)
- Trigonometric Differentiation: \(\displaystyle y' = \frac{1}{3} \bigg( \frac{-\csc x \cot x \sin x - \csc x \cos x}{\sin^2 x} \bigg)\)
- Factor: \(\displaystyle y' = \frac{1}{3} \bigg( \frac{-\csc x (\cot x \sin x + \cos x)}{\sin^2 x} \bigg)\)
- Simplify: \(\displaystyle y' = \frac{-\csc x (\cot x \sin x + \cos x)}{3\sin^2 x}\)
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
