Questions: Find f'(x) and f'(c). Function Value of c f(x) = (x^5 + 4x)(3x^3 + 3x - 5) c = 0 f'(x) = f'(c) =

Find f'(x) and f'(c). Function Value of c f(x) = (x^5 + 4x)(3x^3 + 3x - 5) c = 0 f'(x) = f'(c) =
Transcript text: Find $f^{\prime}(x)$ and $f^{\prime}(c)$. \[ \begin{array}{l} \text { Function } \\ \text { Value of } c \\ f(x)=\left(x^{5}+4 x\right)\left(3 x^{3}+3 x-5\right) \\ c=0 \\ f^{\prime}(x)= \\ f^{\prime}(c)= \end{array} \]
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Solution

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Solution Steps

To find the derivative \( f^{\prime}(x) \) of the given function \( f(x) = (x^5 + 4x)(3x^3 + 3x - 5) \), we will use the product rule. The product rule states that if you have a function \( f(x) = u(x) \cdot v(x) \), then the derivative \( f^{\prime}(x) = u^{\prime}(x) \cdot v(x) + u(x) \cdot v^{\prime}(x) \). After finding \( f^{\prime}(x) \), we will evaluate it at \( c = 0 \) to find \( f^{\prime}(c) \).

Step 1: Define the Function and Apply the Product Rule

Given the function \( f(x) = (x^5 + 4x)(3x^3 + 3x - 5) \), we need to find its derivative \( f^{\prime}(x) \). We apply the product rule, which states that if \( f(x) = u(x) \cdot v(x) \), then \( f^{\prime}(x) = u^{\prime}(x) \cdot v(x) + u(x) \cdot v^{\prime}(x) \).

Step 2: Differentiate Each Component

Let \( u(x) = x^5 + 4x \) and \( v(x) = 3x^3 + 3x - 5 \).

  • The derivative of \( u(x) \) is \( u^{\prime}(x) = 5x^4 + 4 \).
  • The derivative of \( v(x) \) is \( v^{\prime}(x) = 9x^2 + 3 \).
Step 3: Apply the Product Rule

Using the product rule, we find:

\[ f^{\prime}(x) = (5x^4 + 4)(3x^3 + 3x - 5) + (x^5 + 4x)(9x^2 + 3) \]

Step 4: Evaluate the Derivative at \( c = 0 \)

Substitute \( x = 0 \) into \( f^{\prime}(x) \) to find \( f^{\prime}(c) \):

\[ f^{\prime}(0) = (5 \cdot 0^4 + 4)(3 \cdot 0^3 + 3 \cdot 0 - 5) + (0^5 + 4 \cdot 0)(9 \cdot 0^2 + 3) = -20 \]

Final Answer

The derivative of the function is:

\[ f^{\prime}(x) = (9x^2 + 3)(x^5 + 4x) + (5x^4 + 4)(3x^3 + 3x - 5) \]

The value of the derivative at \( c = 0 \) is:

\[ f^{\prime}(0) = \boxed{-20} \]

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