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:
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"