Question

attempt 1: 10 attempts remaining. a. find the derivative of ( g(x) = (7x - 1)(3x + 7) ). ( g(x) = ) b. what is the slope at ( x = -2 )? slope at ( x = -2 ): submit answer next item

Explanation:

Part a: Find the derivative of \( g(x) = (7x - 1)(3x + 7) \)

Step 1: Use the product rule

The product rule states that if \( g(x) = u(x)v(x) \), then \( g'(x) = u'(x)v(x) + u(x)v'(x) \). Let \( u(x) = 7x - 1 \) and \( v(x) = 3x + 7 \).
First, find \( u'(x) \) and \( v'(x) \).
For \( u(x) = 7x - 1 \), the derivative \( u'(x) = 7 \) (since the derivative of \( 7x \) is \( 7 \) and the derivative of a constant \( -1 \) is \( 0 \)).
For \( v(x) = 3x + 7 \), the derivative \( v'(x) = 3 \) (since the derivative of \( 3x \) is \( 3 \) and the derivative of a constant \( 7 \) is \( 0 \)).

Step 2: Apply the product rule

Now, substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the product rule formula:
\[

$$\begin{align*} g'(x) &= u'(x)v(x) + u(x)v'(x) \\ &= 7(3x + 7) + (7x - 1)(3) \end{align*}$$

\]

Step 3: Simplify the expression

First, expand both terms:

• \( 7(3x + 7) = 21x + 49 \)
• \( (7x - 1)(3) = 21x - 3 \)

Now, add the two expanded terms together:
\[

$$\begin{align*} g'(x) &= (21x + 49) + (21x - 3) \\ &= 21x + 21x + 49 - 3 \\ &= 42x + 46 \end{align*}$$

\]

Step 1: Recall the relationship between derivative and slope

The slope of the function \( g(x) \) at a point \( x = a \) is given by the value of the derivative \( g'(a) \). So, we need to evaluate \( g'(-2) \) using the derivative we found in part a, which is \( g'(x) = 42x + 46 \).

Step 2: Substitute \( x = -2 \) into \( g'(x) \)

Substitute \( x = -2 \) into \( g'(x) = 42x + 46 \):
\[

$$\begin{align*} g'(-2) &= 42(-2) + 46 \\ &= -84 + 46 \end{align*}$$

\]

Step 3: Calculate the result

\[
g'(-2) = -38
\]

Answer:

(for part a):
\( g'(x) = 42x + 46 \)

Part b: What is the slope at \( x = -2 \)?