Question

See photo for questions.

Answer 1

1. "a"
`u'=6x`

2. "d"
`v'=15x^2`

3. "b"
`y'=75x^4+30x^2+6x`

Explanation:

General outline:

1. For parts 1 & 2, apply power rule
2. For part 3, apply product rule

Part 1.

Given
`u=3x^2+2`, find
`(du)/(dx) \text{ or } u'`.

`u=3x^2+2`

Apply a derivative to both sides...

`u'=(3x^2+2)'`

Derivatives of a sum are the sum of derivatives...

`u'=(3x^2)'+(2)'`

Scalars factor out of derivatives...

`u'=3(x^2)'+(2)'`

Apply power rule for derivatives (decrease power by 1; mutliply old power as a factor to the coefficient); Derivative of a constant is zero...

`u'=3(2x)+0`

Simplify...

`u'=6x`

So, option "a"

Part 2.

Given
`v=5x^3+1`, find
`(dv)/(dx) \text{ or } v'`.

`v=5x^3+1`

Apply a derivative to both sides...

`v'=(5x^3+1)'`

Derivatives of a sum are the sum of derivatives...

`v'=(5x^3)'+(1)'`

Scalars factor out of derivatives...

`v'=5(x^3)'+(1)'`

Apply power rule for derivatives (decrease power by 1; mutliply old power as a factor to the coefficient); Derivative of a constant is zero...

`v'=5(3x^2)+0`

Simplify...

`v'=15x^2`

So, option "d"

Part 3.

Given
`y=(3x^2+2)(5x^3+1)`

`\text{Then if } u=3x^2+2 \text{ and } v=5x^3+1, y=u*v`

To find
`(dy)/(dx) \text{ or } y'`, recall the product rule:
`y'=uv'+u'v`

`y'=uv'+u'v`

Substituting the expressions found from above...

`y'=(3x^2+2)(15x^2)+(6x)(5x^3+1)`

Apply the distributive property...

`y'=(45x^4+30x^2)+(30x^4+6x)`

Use the associative and commutative property of addition to combine like terms, and rewrite in descending order:

`y'=75x^4+30x^2+6x`

So, option "b"