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Y^4+5x=21 second derivative

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Answer:


\displaystyle y'' = (-75)/(16y^7)

General Formulas and Concepts:

Calculus

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:
\displaystyle (d)/(dx) [cf(x)] = c \cdot f'(x)

Derivative Property [Addition/Subtraction]:
\displaystyle (d)/(dx)[f(x) + g(x)] = (d)/(dx)[f(x)] + (d)/(dx)[g(x)]

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
\displaystyle (d)/(dx)[f(g(x))] =f'(g(x)) \cdot g'(x)

Implicit Differentiation

Explanation:

Step 1: Define

Identify


\displaystyle y^4 + 5x = 21

Step 2: Find 1st Derivative

  1. Differentiate [Basic Power Rule, Chain Rule, Derivative Properties]:
    \displaystyle 4y^3y' + 5 = 0
  2. Isolate y' term:
    \displaystyle 4y^3y' = -5
  3. Isolate y':
    \displaystyle y' = (-5)/(4y^3)

Step 3: Find 2nd Derivative

  1. Rewrite:
    \displaystyle y' = (-5)/(4)y^(-3)
  2. Differentiation [Basic Power Rule, Chain Rule, Derivative Properties]:
    \displaystyle y'' = (15)/(4)y^(-4)y'
  3. Rewrite:
    \displaystyle y'' = (15)/(4y^4)y'
  4. Substitute in y':
    \displaystyle y'' = (15)/(4y^4) \bigg( (-5)/(4y^3) \bigg)
  5. Simplify:
    \displaystyle y'' = (-75)/(16y^7)

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

Unit: Differentiation

User Paul Morie
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