let f(x) = ((x^3)-4)^.5    and let g(x) = 3e^4x       What is the derivitive of g(f(x))?

We are tasked with finding the derivative of the composite function \( g(f(x)) \), where:

- \( f(x) = \sqrt{x^3 - 4} = (x^3 - 4)^{1/2} \)
- \( g(x) = 3e^{4x} \)

The derivative of a composite function can be calculated using the **chain rule**, which states that:

\[
\frac{d}{dx}[g(f(x))] = g'(f(x)) \cdot f'(x)
\]

### Step 1: Find \( f'(x) \)

Starting with \( f(x) = (x^3 - 4)^{1/2} \), we apply the chain rule to find its derivative:

\[
f'(x) = \frac{1}{2}(x^3 - 4)^{-1/2} \cdot \frac{d}{dx}(x^3 - 4)
\]
\[
f'(x) = \frac{1}{2}(x^3 - 4)^{-1/2} \cdot 3x^2
\]
\[
f'(x) = \frac{3x^2}{2\sqrt{x^3 - 4}}
\]

### Step 2: Find \( g'(x) \)

Next, we differentiate \( g(x) = 3e^{4x} \):

\[
g'(x) = 3 \cdot 4e^{4x} = 12e^{4x}
\]

### Step 3: Apply the Chain Rule

Now we use the chain rule to find the derivative of \( g(f(x)) \):

\[
\frac{d}{dx}[g(f(x))] = g'(f(x)) \cdot f'(x)
\]
\[
\frac{d}{dx}[g(f(x))] = 12e^{4f(x)} \cdot \frac{3x^2}{2\sqrt{x^3 - 4}}
\]
\[
\frac{d}{dx}[g(f(x))] = \frac{36x^2 e^{4\sqrt{x^3 - 4}}}{2\sqrt{x^3 - 4}}
\]
\[
\frac{d}{dx}[g(f(x))] = \frac{18x^2 e^{4\sqrt{x^3 - 4}}}{\sqrt{x^3 - 4}}
\]

This is the derivative of \( g(f(x)) \).

Would you like further explanation or more details on any part of this solution?

Here are 5 related questions:
1. How would you apply the chain rule to more complex functions?
2. Can the product rule and chain rule be applied together for certain derivatives?
3. What is the significance of the exponential function in calculus?
4. How do derivatives of radical functions differ from other functions?
5. How can we simplify composite function derivatives in real-world applications?

**Tip:** Always apply the chain rule carefully by working from the outside function to the inside one.

let f(x) = ((x^3)-4)^.5 and let g(x) = 3e^4x What is the derivative of g(f(x))?

Find the derivative of the composite function g(f(x)), where f(x) = √(x^3 - 4) and g(x) = 3e^(4x). Learn how to apply the chain rule and differentiate complex functions step-by-step. Ideal for Grades 11-12 calculus students.

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