Differentiate the given function.
y equals x left parenthesis x squared plus 4 right parenthesis cubedy=xx2+43
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Part 1
StartFraction d Over dx EndFraction left bracket x left parenthesis x squared plus 4 right parenthesis cubed right bracket
d
dxxx2+43equals=

To differentiate the given function \( y = x (x^2 + 4)^3 \), we'll use the **product rule** and **chain rule**.

### Step 1: Apply the Product Rule
The product rule states:
\[
\frac{d}{dx} [u \cdot v] = u' \cdot v + u \cdot v'
\]
Here, \( u = x \) and \( v = (x^2 + 4)^3 \). We'll differentiate each part.

- For \( u = x \), we have:
\[
u' = \frac{d}{dx}(x) = 1
\]

- For \( v = (x^2 + 4)^3 \), we apply the **chain rule**. Let \( g(x) = x^2 + 4 \), so:
\[
v = g(x)^3
\]
The derivative of \( g(x)^3 \) is:
\[
v' = 3g(x)^2 \cdot g'(x)
\]
where \( g'(x) = \frac{d}{dx}(x^2 + 4) = 2x \).

Thus,
\[
v' = 3(x^2 + 4)^2 \cdot 2x = 6x(x^2 + 4)^2
\]

### Step 2: Apply the Product Rule
Now, using the product rule:
\[
\frac{d}{dx} \left[ x(x^2 + 4)^3 \right] = 1 \cdot (x^2 + 4)^3 + x \cdot 6x(x^2 + 4)^2
\]

### Step 3: Simplify the Expression
\[
\frac{d}{dx} \left[ x(x^2 + 4)^3 \right] = (x^2 + 4)^3 + 6x^2(x^2 + 4)^2
\]

This is the derivative of the given function.

Would you like more details on any part of this? Here are a few related questions:

1. How can we further simplify the final derivative?
2. What is the general formula for the product rule?
3. How does the chain rule apply in different contexts?
4. What would the second derivative of this function look like?
5. How would we find the critical points of this function?

**Tip:** When dealing with products of functions, always consider the product rule first before applying other differentiation techniques.

Differentiate the given function. y equals x left parenthesis x squared plus 4 right parenthesis cubedy=xx2+43

Learn how to differentiate the function y = x(x^2 + 4)^3 using both the product rule and chain rule. This detailed step-by-step guide covers key concepts in calculus such as differentiation, product rule, and chain rule. Suitable for students in Grades 11-12 or Calculus I.

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