Math Problem Statement

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

Solution

To differentiate the given function y=x(x2+4)3y = x (x^2 + 4)^3, we'll use the product rule and chain rule.

Step 1: Apply the Product Rule

The product rule states: ddx[uv]=uv+uv\frac{d}{dx} [u \cdot v] = u' \cdot v + u \cdot v' Here, u=xu = x and v=(x2+4)3v = (x^2 + 4)^3. We'll differentiate each part.

  • For u=xu = x, we have: u=ddx(x)=1u' = \frac{d}{dx}(x) = 1

  • For v=(x2+4)3v = (x^2 + 4)^3, we apply the chain rule. Let g(x)=x2+4g(x) = x^2 + 4, so: v=g(x)3v = g(x)^3 The derivative of g(x)3g(x)^3 is: v=3g(x)2g(x)v' = 3g(x)^2 \cdot g'(x) where g(x)=ddx(x2+4)=2xg'(x) = \frac{d}{dx}(x^2 + 4) = 2x.

Thus, v=3(x2+4)22x=6x(x2+4)2v' = 3(x^2 + 4)^2 \cdot 2x = 6x(x^2 + 4)^2

Step 2: Apply the Product Rule

Now, using the product rule: ddx[x(x2+4)3]=1(x2+4)3+x6x(x2+4)2\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

ddx[x(x2+4)3]=(x2+4)3+6x2(x2+4)2\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.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Product Rule
Chain Rule

Formulas

Product Rule: d/dx[u * v] = u' * v + u * v'
Chain Rule: d/dx[f(g(x))] = f'(g(x)) * g'(x)

Theorems

Product Rule
Chain Rule

Suitable Grade Level

Grades 11-12 or Calculus I