Question

Can someone help me with this please

Answer 1

`(d^(2)y)/(dx^(2)) = 180x.(x^(3) +8)^(4)`

Explanation:

This question requires use of the chain rule:

`y = (x^(3) + 8)^(6)`

Let:

`u = x^(3) + 8`

Then:

`y = u^(6)`

And:

`(dy)/(dx) = (dy)/(du). (du)/(dx)\\\\(d^(2)y)/(dx^(2) ) = (d^(2)y)/(du^(2)). (d^(2)u)/(dx^(2))\\\\(dy)/(du) = 6u^(5)\\\\(d^(2)y)/(du^(2)) = 30u^(4)\\\\(du)/(dx) = 3x^(2)\\\\(d^(2)u)/(dx^(2)) = 6x`

So:

`(dy)/(dx) = 6u^(5).3x^(2)\\\\(dy)/(dx) = 18u^(5)x^(2)\\\\`

And:

`(d^(2)y)/(dx^(2)) = 30u^(4).6x\\\\(d^(2)y)/(dx^(2)) = 180u^(4)x`

Finally, substitute u for its equivalent expression in x:

`(d^(2)y)/(dx^(2)) = 180x.(x^(3) +8)^(4)`

Answer 2

y'' = 18x(x³ + 8)⁴(17x³ + 16)

General Formulas and Concepts:

Calculus

Derivative Notation dy/dx

Derivative of a constant is 0.

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

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

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

Explanation:

Step 1: Define

y = (x³ + 8)⁶

Step 2: Find 1st Derivative (dy/dx)

1. Chain Rule [Basic Power]: y' = 6(x³ + 8)⁶⁻¹ · (3x³⁻¹ + 0)
2. Simplify: y' = 6(x³ + 8)⁵ · 3x²
3. Multiply: y' = 18x²(x³ + 8)⁵

Step 3: Find 2nd Derivative (d²y/dx²)

1. Product Rule [Chain/Basic Power]: y'' = 2 · 18x²⁻¹ · (x³ + 8)⁵ + 18x² · 5(x³ + 8)⁵⁻¹ · (3x³⁻¹ + 0)
2. Simplify: y'' = 36x(x³ + 8)⁵ + 90x²(x³ + 8)⁴ · 3x²
3. Multiply: y'' = 36x(x³ + 8)⁵ + 270x⁴(x³ + 8)⁴
4. Factor: y'' = 6x(x³ + 8)⁴[6(x³ + 8) + 45x³]
5. Distribute: y'' = 6x(x³ + 8)⁴[6x³ + 48 + 45x³]
6. Combine like terms: y'' = 6x(x³ + 8)⁴[51x³ + 48]
7. Factor: y'' = 6x(x³ + 8)⁴ · 3(17x³ + 16)
8. Multiply: y'' = 18x(x³ + 8)⁴(17x³ + 16)