Question

Can someone help me with this please

Answer 1

Answer:

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: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Product Rule: $\frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step 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)

Answer 2

Answer:

$\frac{d^{2}y}{dx^{2}} = 180x.(x^{3} +8)^{4}$

Step-by-step explanation:

This question requires use of the chain rule:

$y = (x^{3} + 8)^{6}$

Let:

$u = x^{3} + 8$

Then:

$y = u^{6}$

And:

$\frac{dy}{dx} = \frac{dy}{du}. \frac{du}{dx}\\\\\frac{d^{2}y}{dx^{2} } = \frac{d^{2}y}{du^{2}}. \frac{d^{2}u}{dx^{2}}\\\\\frac{dy}{du} = 6u^{5}\\\\\frac{d^{2}y}{du^{2}} = 30u^{4}\\\\\frac{du}{dx} = 3x^{2}\\\\\frac{d^{2}u}{dx^{2}} = 6x$

So:

$\frac{dy}{dx} = 6u^{5}.3x^{2}\\\\\frac{dy}{dx} = 18u^{5}x^{2}$

And:

$\frac{d^{2}y}{dx^{2}} = 30u^{4}.6x\\\\\frac{d^{2}y}{dx^{2}} = 180u^{4}x$

Finally, substitute u for its equivalent expression in x:

$\frac{d^{2}y}{dx^{2}} = 180x.(x^{3} +8)^{4}$