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Find the derivative of cscx/3sinx

User TauWich
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Answer:


\displaystyle (dy)/(dx) = (-\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 (d)/(dx) [cf(x)] = c \cdot f'(x)

Derivative Rule [Quotient Rule]:
\displaystyle (d)/(dx) [(f(x))/(g(x)) ]=(g(x)f'(x)-g'(x)f(x))/(g^2(x))

Explanation:

Step 1: Define

Identify


\displaystyle y = (\csc x)/(3\sin x)

Step 2: Differentiate

  1. Derivative Property [Multiplied Constant]:
    \displaystyle y' = (1)/(3) (d)/(dx)[(\csc x)/(\sin x)]
  2. Derivative Rule [Quotient Rule]:
    \displaystyle y' = (1)/(3) \bigg( ((\csc x)' \sin x - \csc x (\sin x)')/(\sin^2 x) \bigg)
  3. Trigonometric Differentiation:
    \displaystyle y' = (1)/(3) \bigg( (-\csc x \cot x \sin x - \csc x \cos x)/(\sin^2 x) \bigg)
  4. Factor:
    \displaystyle y' = (1)/(3) \bigg( (-\csc x (\cot x \sin x + \cos x))/(\sin^2 x) \bigg)
  5. Simplify:
    \displaystyle y' = (-\csc x (\cot x \sin x + \cos x))/(3\sin^2 x)

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

User Zully
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