ANSWERS
• f'(x) = 18x⁵ + 32x³ - 6x² + 8x - 4
,
• f'(c) = -4
EXPLANATION
To find the derivative of this function we can apply the product rule,
[tex]\begin{gathered} f(x)=g(x)\cdot h(x) \\ f^{\prime}(x)=g^{\prime}(x)\cdot h(x)+g(x)\cdot h^{\prime}(x) \end{gathered}[/tex]
In this case, the functions are,
• g(x) = x³ + 2x
,
• h(x) = 3x³ + 2x - 2
The derivative of g(x) is,
[tex]g^{\prime}(x)=3x^2+2[/tex]
The derivative of h(x) is,
[tex]h^{\prime}(x)=9x^2+2[/tex]
So, the derivative of f(x) is,
[tex]f^{\prime}(x)=(3x^2+2)(3x^3+2x-2)+(x^3+2x)(9x^2+2)[/tex]
Let's simplify the answer. Multiply the two parts in the first term,
[tex]\begin{gathered} f^{\prime}(x)=(3x^2\cdot3x^3+3x^2\cdot2x-3x^2\cdot2+2\cdot3x^3+2\cdot2x-2\cdot2)+(x^3+2x)(9x^2+2) \\ f^{\prime}(x)=(9x^5+6x^3-6x^2+6x^3+4x-4)+(x^3+2x)(9x^2+2) \end{gathered}[/tex]
Add like terms,
[tex]f^{\prime}(x)=(9x^5+12x^3-6x^2+4x-4)+(x^3+2x)(9x^2+2)[/tex]
Do the same for the second term,
[tex]\begin{gathered} f^{\prime}(x)=(9x^5+12x^3-6x^2+4x-4)+(x^3\cdot9x^2+x^3\cdot2+2x\cdot9x^2+2x\cdot2) \\ f^{\prime}(x)=(9x^5+12x^3-6x^2+4x-4)+(9x^5^{}+2x^3+18x^3+4x) \end{gathered}[/tex]
Add like terms,
[tex]\begin{gathered} f^{\prime}(x)=(9x^5+9x^5)+(12x^3+2x^3+18x^3)-6x^2+(4x+4x)-4 \\ f^{\prime}(x)=18x^5+32x^3-6x^2+8x-4 \end{gathered}[/tex]
Hence, the derivative of f(x) is f'(x) = 18x⁵ + 32x³ - 6x² + 8x - 4.
Now, we have to find the value of f'(0). Note that all the terms of f'(x) except for the last one contain x, so they all are 0 except for the last term. Hence, the value of f'(0) is -4
