The given problem involves identifying the composite function $f(g(x))$, where the function is given as:

$y = \sqrt[3]{1 + 4x}$

Step 1: Identify the inner and outer functions

We need to express the function in the form $y = f(g(x))$.

• The inner function $g(x)$ is the expression inside the cube root: $g(x) = 1 + 4x$

• The outer function $f(u)$ is the cube root applied to the inner function: $f(u) = \sqrt[3]{u}$

Thus, we can express the composite function as $(g(x), f(u)) = (1 + 4x, \sqrt[3]{u})$.

Step 2: Find the derivative $\frac{dy}{dx}$

Now, to find $\frac{dy}{dx}$, we apply the chain rule. The chain rule states: $\frac{dy}{dx} = \frac{df}{du} \cdot \frac{dg}{dx}$

• First, find $\frac{df}{du}$ for the outer function: $f(u) = u^{1/3} \quad \Rightarrow \quad \frac{df}{du} = \frac{1}{3} u^{-2/3}$

• Next, find $\frac{dg}{dx}$ for the inner function: $g(x) = 1 + 4x \quad \Rightarrow \quad \frac{dg}{dx} = 4$

Now, substitute $u = g(x) = 1 + 4x$ into $\frac{df}{du}$: $\frac{df}{du} = \frac{1}{3} (1 + 4x)^{-2/3}$

Finally, apply the chain rule: $\frac{dy}{dx} = \frac{1}{3} (1 + 4x)^{-2/3} \cdot 4 = \frac{4}{3} (1 + 4x)^{-2/3}$

Final Answer:

• Inner function: $g(x) = 1 + 4x$
• Outer function: $f(u) = \sqrt[3]{u}$
• Derivative: $\frac{dy}{dx} = \frac{4}{3} (1 + 4x)^{-2/3}$

Would you like further details or have any other questions?

Here are 5 related questions to expand your understanding:

1. How would the process change if the outer function were a square root instead of a cube root?
2. Can you explain how the chain rule works for composite functions involving trigonometric functions?
3. What if the inner function were more complex, such as $g(x) = x^2 + 4x$? How would that affect the derivative?
4. How does the chain rule extend to functions with more than two compositions?
5. What is the geometric interpretation of the derivative in the context of composite functions?

Tip: When applying the chain rule, always differentiate the outer function first and then multiply by the derivative of the inner function.