Derivative Product Rule 1E0B54

1. **State the problem:** We need to find the derivative $g'(2)$ where $g(x) = (4x^3 - 2x + 6) f(x)$ and $f(x)$ is a function with a known tangent line at $x=2$. 2. **Recall the product rule:** For two functions $u(x)$ and $v(x)$, the derivative of their product is $$g'(x) = u'(x)v(x) + u(x)v'(x).$$ Here, let $$u(x) = 4x^3 - 2x + 6$$ and $$v(x) = f(x).$$ 3. **Compute $u'(x)$:** $$u'(x) = \frac{d}{dx}(4x^3 - 2x + 6) = 12x^2 - 2.$$ 4. **Evaluate $u(2)$ and $u'(2)$:** $$u(2) = 4(2)^3 - 2(2) + 6 = 4 \times 8 - 4 + 6 = 32 - 4 + 6 = 34,$$ $$u'(2) = 12(2)^2 - 2 = 12 \times 4 - 2 = 48 - 2 = 46.$$ 5. **Use the tangent line to find $f(2)$ and $f'(2)$:** The tangent line touches the graph at $(2, 2)$, so $$f(2) = 2.$$ The slope of the tangent line at $x=2$ is the derivative $f'(2)$. Since the tangent line is dashed and touches at $(2,2)$, and the parabola is upward-opening, the slope is positive. From the graph description, the slope is the slope of the tangent line at $x=2$. We can approximate or assume the slope is the derivative at that point. Since no explicit slope is given, assume the slope is the slope of the tangent line shown. Let's denote it as $f'(2) = m$. 6. **Calculate $g'(2)$ using the product rule:** $$g'(2) = u'(2) f(2) + u(2) f'(2) = 46 \times 2 + 34 \times f'(2) = 92 + 34 f'(2).$$ 7. **Find $f'(2)$ from the tangent line:** The tangent line at $x=2$ passes through $(2,2)$ and has slope $f'(2)$. Since the parabola is upward-opening and the tangent line is dashed, the slope is positive. Without explicit slope, we cannot numerically evaluate $g'(2)$ further. **Final answer:** $$g'(2) = 92 + 34 f'(2).$$ If the slope $f'(2)$ is known from the graph, substitute it here to get the numerical value.