Question

Find the derivatives of the function f for n = 1, 2, 3, and 4.

Answer 1

ANSWERS

• f'(x) = 18x⁵ + 32x³ - 6x² + 8x - 4

,

• f'(c) = -4

Step-by-step explanation

To find the derivative of this function we can apply the product rule,

`\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}`

In this case, the functions are,

• g(x) = x³ + 2x

,

• h(x) = 3x³ + 2x - 2

The derivative of g(x) is,

`g^(\prime)(x)=3x^2+2`

The derivative of h(x) is,

`h^(\prime)(x)=9x^2+2`

So, the derivative of f(x) is,

`f^(\prime)(x)=(3x^2+2)(3x^3+2x-2)+(x^3+2x)(9x^2+2)`

Let's simplify the answer. Multiply the two parts in the first term,

`\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}`

Add like terms,

`f^(\prime)(x)=(9x^5+12x^3-6x^2+4x-4)+(x^3+2x)(9x^2+2)`

Do the same for the second term,

`\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}`

Add like terms,

`\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}`

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