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) =

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

Solution

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 $.