Question

If -y-2x^3=Y^2 then find D^2y/dx^2 at the point (-1,-2) in simplest form

^^ the main problem i am having is when I have to use the quotient rule

Answer 1

`(d^2y)/(dx^2) = (-4)/(3)`

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Algebra I

• Factoring

Calculus

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

Product Rule:
`(d)/(dx) [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)`

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

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

Explanation:

Step 1: Define

-y - 2x³ = y²

Rate of change of tangent line at point (-1, -2)

Step 2: Differentiate Pt. 1

Find 1st Derivative

1. Implicit Differentiation [Basic Power Rule]:
   `-y'-6x^2=2yy'`
2. [Algebra] Isolate y' terms:
   `-6x^2=2yy'+y'`
3. [Algebra] Factor y':
   `-6x^2=y'(2y+1)`
4. [Algebra] Isolate y':
   `(-6x^2)/((2y+1))=y'`
5. [Algebra] Rewrite:
   `y' = (-6x^2)/((2y+1))`

Step 3: Differentiate Pt. 2

Find 2nd Derivative

1. Differentiate [Quotient Rule/Basic Power Rule]:
   `y'' = (-12x(2y+1)+6x^2(2y'))/((2y+1)^2)`
2. [Derivative] Simplify:
   `y'' = (-24xy-12x+12x^2y')/((2y+1)^2)`
3. [Derivative] Back-Substitute y':
   `y'' = (-24xy-12x+12x^2((-6x^2)/(2y+1) ))/((2y+1)^2)`
4. [Derivative] Simplify:
   `y'' = (-24xy-12x-(72x^4)/(2y+1) )/((2y+1)^2)`

Step 4: Find Slope at Given Point

1. [Algebra] Substitute in x and y:
   `y''(-1,-2) = (-24(-1)(-2)-12(-1)-(72(-1)^4)/(2(-2)+1) )/((2(-2)+1)^2)`
2. [Pre-Algebra] Exponents:
   `y''(-1,-2) = (-24(-1)(-2)-12(-1)-(72(1))/(2(-2)+1) )/((2(-2)+1)^2)`
3. [Pre-Algebra] Multiply:
   `y''(-1,-2) = (-48+12-(72)/(-4+1) )/((-4+1)^2)`
4. [Pre-Algebra] Add:
   `y''(-1,-2) = (-36-(72)/(-3) )/((-3)^2)`
5. [Pre-Algebra] Exponents:
   `y''(-1,-2) = (-36-(72)/(-3) )/(9)`
6. [Pre-Algebra] Divide:
   `y''(-1,-2) = (-36+24 )/(9)`
7. [Pre-Algebra] Add:
   `y''(-1,-2) = (-12)/(9)`
8. [Pre-Algebra] Simplify:
   `y''(-1,-2) = (-4)/(3)`