Answer :
Answer:
[tex]\displaystyle y'' = \frac{-75}{16y^7}[/tex]
General Formulas and Concepts:
Calculus
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]: [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]
Derivative Property [Addition/Subtraction]: [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]
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]
Implicit Differentiation
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle y^4 + 5x = 21[/tex]
Step 2: Find 1st Derivative
- Differentiate [Basic Power Rule, Chain Rule, Derivative Properties]: [tex]\displaystyle 4y^3y' + 5 = 0[/tex]
- Isolate y' term: [tex]\displaystyle 4y^3y' = -5[/tex]
- Isolate y': [tex]\displaystyle y' = \frac{-5}{4y^3}[/tex]
Step 3: Find 2nd Derivative
- Rewrite: [tex]\displaystyle y' = \frac{-5}{4}y^{-3}[/tex]
- Differentiation [Basic Power Rule, Chain Rule, Derivative Properties]: [tex]\displaystyle y'' = \frac{15}{4}y^{-4}y'[/tex]
- Rewrite: [tex]\displaystyle y'' = \frac{15}{4y^4}y'[/tex]
- Substitute in y': [tex]\displaystyle y'' = \frac{15}{4y^4} \bigg( \frac{-5}{4y^3} \bigg)[/tex]
- Simplify: [tex]\displaystyle y'' = \frac{-75}{16y^7}[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation