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Sagot :

Answer:

1.  "a"  [tex]u'=6x[/tex]

2.  "d"  [tex]v'=15x^2[/tex]

3.  "b"  [tex]y'=75x^4+30x^2+6x[/tex]

Step-by-step explanation:

General outline:

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

Part 1.

Given [tex]u=3x^2+2[/tex], find [tex]\frac{du}{dx} \text{ or } u'[/tex].

[tex]u=3x^2+2[/tex]

Apply a derivative to both sides...

[tex]u'=(3x^2+2)'[/tex]

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

[tex]u'=(3x^2)'+(2)'[/tex]

Scalars factor out of derivatives...

[tex]u'=3(x^2)'+(2)'[/tex]

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

[tex]u'=3(2x)+0[/tex]

Simplify...

[tex]u'=6x[/tex]

So, option "a"

Part 2.

Given [tex]v=5x^3+1[/tex], find [tex]\frac{dv}{dx} \text{ or } v'[/tex].

[tex]v=5x^3+1[/tex]

Apply a derivative to both sides...

[tex]v'=(5x^3+1)'[/tex]

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

[tex]v'=(5x^3)'+(1)'[/tex]

Scalars factor out of derivatives...

[tex]v'=5(x^3)'+(1)'[/tex]

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

[tex]v'=5(3x^2)+0[/tex]

Simplify...

[tex]v'=15x^2[/tex]

So, option "d"

Part 3.

Given [tex]y=(3x^2+2)(5x^3+1)[/tex]

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

To find [tex]\frac{dy}{dx} \text{ or } y'[/tex], recall the product rule:  [tex]y'=uv'+u'v[/tex]

[tex]y'=uv'+u'v[/tex]

Substituting the expressions found from above...

[tex]y'=(3x^2+2)(15x^2)+(6x)(5x^3+1)[/tex]

Apply the distributive property...

[tex]y'=(45x^4+30x^2)+(30x^4+6x)[/tex]

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

[tex]y'=75x^4+30x^2+6x[/tex]

So, option "b"

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