Math Problem Statement

Differentiate the function.

y equals left parenthesis 6 x Superscript 4 Baseline minus x plus 4 right parenthesis left parenthesis negative x Superscript 5 Baseline plus 9 right parenthesis

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Part 1

y prime equals nothing       

Solution

To differentiate the function y=(6x4x+4)(x5+9)y = (6x^4 - x + 4)(-x^5 + 9), we'll use the product rule. The product rule states that if y=u(x)v(x)y = u(x) \cdot v(x), then:

y=u(x)v(x)+u(x)v(x)y' = u'(x) \cdot v(x) + u(x) \cdot v'(x)

Here, we can identify:

u(x)=6x4x+4u(x) = 6x^4 - x + 4 v(x)=x5+9v(x) = -x^5 + 9

Step 1: Differentiate u(x)u(x) and v(x)v(x)

First, let's find the derivatives of u(x)u(x) and v(x)v(x):

u(x)=ddx(6x4x+4)=24x31u'(x) = \frac{d}{dx}(6x^4 - x + 4) = 24x^3 - 1

v(x)=ddx(x5+9)=5x4v'(x) = \frac{d}{dx}(-x^5 + 9) = -5x^4

Step 2: Apply the Product Rule

Now, apply the product rule:

y=u(x)v(x)+u(x)v(x)y' = u'(x) \cdot v(x) + u(x) \cdot v'(x)

Substitute the derivatives and the original functions:

y=(24x31)(x5+9)+(6x4x+4)(5x4)y' = (24x^3 - 1)(-x^5 + 9) + (6x^4 - x + 4)(-5x^4)

Step 3: Expand the Expression

Now, we'll expand the expression to simplify it:

Expand the first term:

(24x31)(x5+9)=24x3(x5)+24x3(9)1(x5)1(9)(24x^3 - 1)(-x^5 + 9) = 24x^3(-x^5) + 24x^3(9) - 1(-x^5) - 1(9) =24x8+216x3+x59= -24x^8 + 216x^3 + x^5 - 9

Expand the second term:

(6x4x+4)(5x4)=6x4(5x4)x(5x4)+4(5x4)(6x^4 - x + 4)(-5x^4) = 6x^4(-5x^4) - x(-5x^4) + 4(-5x^4) =30x8+5x520x4= -30x^8 + 5x^5 - 20x^4

Step 4: Combine Like Terms

Finally, add the expanded expressions together:

y=(24x8+216x3+x59)+(30x8+5x520x4)y' = (-24x^8 + 216x^3 + x^5 - 9) + (-30x^8 + 5x^5 - 20x^4)

Combine like terms:

y=(24x830x8)+(x5+5x5)20x4+216x39y' = (-24x^8 - 30x^8) + (x^5 + 5x^5) - 20x^4 + 216x^3 - 9 y=54x8+6x520x4+216x39y' = -54x^8 + 6x^5 - 20x^4 + 216x^3 - 9

So, the derivative is:

y=54x8+6x520x4+216x39y' = -54x^8 + 6x^5 - 20x^4 + 216x^3 - 9

Would you like further details or have any questions about these steps?

Here are some related questions to expand your understanding:

  1. How does the product rule change if one of the functions is a constant?
  2. What is the significance of the higher-order terms in the derivative of a product?
  3. How would the derivative change if the original function was divided instead of multiplied?
  4. How does the derivative behave as xx approaches infinity?
  5. How would you apply the quotient rule to a similar problem?

Tip: When expanding polynomials during differentiation, be careful with the signs, especially when negative terms are involved.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Product Rule
Polynomial Functions

Formulas

Product Rule for differentiation

Theorems

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Suitable Grade Level

Advanced High School