Question

Y^4+5x=21 second derivative

Answer 1

`\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