AP Calculus AB Exam Question:

Let $f(x)$ and $g(x)$ be differentiable functions such that $f(x) = 2x^3 - x^2 + 4x$ and $g(x) = e^{3x}$.

a) Find $f'(x)$ and $g'(x)$.

b) Use the chain rule to find $\frac{d}{dx} (f \circ g)(x)$.

c) Evaluate $\frac{d}{dx} (f \circ g)(2)$.

Solution:

a) To find $f'(x)$, we differentiate $f(x)$ term by term:

$f(x) = 2x^3 - x^2 + 4x$

$f'(x) = \frac{d}{dx} (2x^3) - \frac{d}{dx} (x^2) + \frac{d}{dx} (4x)$

Using the power rule, we have:

$f'(x) = 6x^2 - 2x + 4$

Now let's find $g'(x)$ using the chain rule. Since $g(x) = e^{3x}$, we have $\frac{d}{dx} g(x) = \frac{d}{dx} e^{3x}$:

$\frac{d}{dx} g(x) = 3e^{3x}$

b) The chain rule states that if $y = f(g(x))$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

Applying the chain rule to $(f \circ g)(x)$, we have:

$\frac{d}{dx}(f \circ g)(x) = f'(g(x)) \cdot g'(x)$

We already found $f'(x)$ and $g'(x)$ in part a), so substituting those values we get:

$\frac{d}{dx}(f \circ g)(x) = (6g^2(x) - 2g(x) + 4) \cdot g'(x)$

Since $g(x) = e^{3x}$ and $g'(x) = 3e^{3x}$, we can substitute these values:

$\frac{d}{dx}(f \circ g)(x) = (6(e^{3x})^2 - 2(e^{3x}) + 4) \cdot 3e^{3x}$

Simplifying further:

$\frac{d}{dx}(f \circ g)(x) = (6e^{6x} - 2e^{3x} + 4) \cdot 3e^{3x}$

c) To evaluate $\frac{d}{dx}(f \circ g)(2)$, we substitute $x=2$ into the expression for $\frac{d}{dx}(f \circ g)(x)$:

$\frac{d}{dx}(f \circ g)(2) = (6e^{6(2)} - 2e^{3(2)} + 4) \cdot 3e^{3(2)}$

Simplifying further:

$\frac{d}{dx}(f \circ g)(2) = (6e^{12} - 2e^6 + 4) \cdot 3e^6$

Since $e$ is a constant, we can evaluate the expression numerically:

$\frac{d}{dx}(f \circ g)(2) \approx (6 \cdot 162754.7914 - 2 \cdot 403.4288 + 4) \cdot 3 \cdot 403.4288$

$\frac{d}{dx}(f \circ g)(2) \approx 976528.7484$

Therefore, $\frac{d}{dx}(f \circ g)(2) \approx 976528.7484$.

Answer: $\frac{d}{dx}(f \circ g)(2) \approx 976528.7484$