AP Calculus AB Exam Question: Chain Rule

Consider the function defined by $f(x) = (2x^3 - x^2) \cdot \sqrt[3]{(4x-1)^2}$.

a) Find the derivative of $f(x)$ with respect to $x$.

b) Evaluate the derivative at $x = 2$.

Answer:

a) To find the derivative of $f(x)$, we will apply the chain rule. Let's break down the function into two parts, denoted as $u$ and $v$:

$u(x) = 2x^3 - x^2$

$v(x) = (4x-1)^2$

Now, we can write the function $f(x)$ as $f(x) = u(x) \cdot v(x)^{\frac{1}{3}}$.

According to the chain rule, the derivative of $f(x)$ can be calculated using the formula:

$\frac{df}{dx} = \frac{du}{dx} \cdot v(x)^{\frac{1}{3}} + u(x) \cdot \frac{dv}{dx} \cdot \frac{1}{3} \cdot v(x)^{-\frac{2}{3}}$

Let's calculate each part individually:

1. $\frac{du}{dx}$: Since $u(x) = 2x^3 - x^2$, the derivative of $u$ with respect to $x$ is:

$\frac{du}{dx} = 6x^2 - 2x$

2. $\frac{dv}{dx}$: We differentiate $v(x) = (4x-1)^2$ by applying the chain rule:

$\frac{dv}{dx} = 2(4x-1) \cdot 4 = 8(4x-1)$

Now, let's substitute the derivatives we found back into the chain rule formula:

$\frac{df}{dx} = (6x^2 - 2x) \cdot (4x-1)^{\frac{1}{3}} + (2x^3 - x^2) \cdot \frac{8(4x-1)}{3} \cdot (4x-1)^{-\frac{2}{3}}$

Simplifying, we get:

$\frac{df}{dx} = (6x^2 - 2x) \cdot (4x-1)^{\frac{1}{3}} + \frac{16}{3} (2x^3 - x^2) \cdot (4x-1)^{-\frac{2}{3}}$

b) Now, let's evaluate the derivative at $x = 2$. Substituting $x = 2$ into the derivative expression we obtained in part (a):

$\frac{df}{dx} \Bigr|_{x=2} = (6(2)^2 - 2(2)) \cdot (4(2)-1)^{\frac{1}{3}} + \frac{16}{3} (2)^3 - (2)^2) \cdot (4(2)-1)^{-\frac{2}{3}}$

Simplifying the expression:

$\frac{df}{dx} \Bigr|_{x=2} = (24 - 4) \cdot (7)^{\frac{1}{3}} + \frac{16}{3} \cdot (8 - 4) \cdot (7)^{-\frac{2}{3}}$

Using a calculator, we find:

$\frac{df}{dx} \Bigr|_{x=2} = 20(7)^{\frac{1}{3}} + \frac{16}{3} \cdot 4 \cdot (7)^{-\frac{2}{3}}$

Simplifying further:

$\frac{df}{dx} \Bigr|_{x=2} = 20(7)^{\frac{1}{3}} + \frac{64}{3} (7)^{-\frac{2}{3}}$

Hence, the derivative of $f(x)$ at $x = 2$ is $20(7)^{\frac{1}{3}} + \frac{64}{3} (7)^{-\frac{2}{3}}$.