Answer :
To find the third derivative, [tex]\( y^{\prime \prime \prime} \)[/tex], for the function [tex]\( y = 3x^5 + 4x^3 + 2x \)[/tex], we will differentiate the function three times. Here’s the step-by-step process:
1. First Derivative ([tex]\( y' \)[/tex]): Differentiate [tex]\( y = 3x^5 + 4x^3 + 2x \)[/tex] term by term.
- The derivative of [tex]\( 3x^5 \)[/tex] is [tex]\( 15x^4 \)[/tex].
- The derivative of [tex]\( 4x^3 \)[/tex] is [tex]\( 12x^2 \)[/tex].
- The derivative of [tex]\( 2x \)[/tex] is [tex]\( 2 \)[/tex].
Thus,
[tex]\[
y' = 15x^4 + 12x^2 + 2.
\][/tex]
2. Second Derivative ([tex]\( y'' \)[/tex]): Differentiate [tex]\( y' = 15x^4 + 12x^2 + 2 \)[/tex] term by term.
- The derivative of [tex]\( 15x^4 \)[/tex] is [tex]\( 60x^3 \)[/tex].
- The derivative of [tex]\( 12x^2 \)[/tex] is [tex]\( 24x \)[/tex].
- The derivative of [tex]\( 2 \)[/tex] is [tex]\( 0 \)[/tex] (since it's a constant).
Thus,
[tex]\[
y'' = 60x^3 + 24x.
\][/tex]
3. Third Derivative ([tex]\( y''' \)[/tex]): Differentiate [tex]\( y'' = 60x^3 + 24x \)[/tex] term by term.
- The derivative of [tex]\( 60x^3 \)[/tex] is [tex]\( 180x^2 \)[/tex].
- The derivative of [tex]\( 24x \)[/tex] is [tex]\( 24 \)[/tex].
Thus,
[tex]\[
y''' = 180x^2 + 24.
\][/tex]
So, the correct third derivative, [tex]\( y^{\prime \prime \prime} \)[/tex], is [tex]\( 180x^2 + 24 \)[/tex].
The correct answer is option B: [tex]\( 180x^2 + 24 \)[/tex].
1. First Derivative ([tex]\( y' \)[/tex]): Differentiate [tex]\( y = 3x^5 + 4x^3 + 2x \)[/tex] term by term.
- The derivative of [tex]\( 3x^5 \)[/tex] is [tex]\( 15x^4 \)[/tex].
- The derivative of [tex]\( 4x^3 \)[/tex] is [tex]\( 12x^2 \)[/tex].
- The derivative of [tex]\( 2x \)[/tex] is [tex]\( 2 \)[/tex].
Thus,
[tex]\[
y' = 15x^4 + 12x^2 + 2.
\][/tex]
2. Second Derivative ([tex]\( y'' \)[/tex]): Differentiate [tex]\( y' = 15x^4 + 12x^2 + 2 \)[/tex] term by term.
- The derivative of [tex]\( 15x^4 \)[/tex] is [tex]\( 60x^3 \)[/tex].
- The derivative of [tex]\( 12x^2 \)[/tex] is [tex]\( 24x \)[/tex].
- The derivative of [tex]\( 2 \)[/tex] is [tex]\( 0 \)[/tex] (since it's a constant).
Thus,
[tex]\[
y'' = 60x^3 + 24x.
\][/tex]
3. Third Derivative ([tex]\( y''' \)[/tex]): Differentiate [tex]\( y'' = 60x^3 + 24x \)[/tex] term by term.
- The derivative of [tex]\( 60x^3 \)[/tex] is [tex]\( 180x^2 \)[/tex].
- The derivative of [tex]\( 24x \)[/tex] is [tex]\( 24 \)[/tex].
Thus,
[tex]\[
y''' = 180x^2 + 24.
\][/tex]
So, the correct third derivative, [tex]\( y^{\prime \prime \prime} \)[/tex], is [tex]\( 180x^2 + 24 \)[/tex].
The correct answer is option B: [tex]\( 180x^2 + 24 \)[/tex].
