Math Problem Statement

Use the values in the chart to evaluate the derivatives of the combinations of functions at given x-values. Show your thinking/work on each one.

Solution

Let's analyze and solve each part of this problem using the provided information from the chart, graphs, and equations.

Given Information:

  1. Functions:
    • j(x)=x3j(x) = x^3
    • k(x)=5xk(x) = 5x
    • n(x)=1xn(x) = \frac{1}{x}
  2. Derivatives provided in the chart for values of f(x)f(x), g(x)g(x), h(x)h(x), and p(x)p(x) along with their derivatives f(x)f'(x), g(x)g'(x), h(x)h'(x), and p(x)p'(x) at specific xx values.

Solution Outline

I’ll walk through each part step-by-step to provide answers and explanations.

(a) Find ddx[m(x)p(x)]x=5\frac{d}{dx} [m(x) p(x)] \Big|_{x=5}

Using the product rule: ddx[m(x)p(x)]=m(x)p(x)+m(x)p(x)\frac{d}{dx} [m(x) p(x)] = m(x) p'(x) + m'(x) p(x) Evaluate at x=5x = 5:

  • From the table, m(5)=2m(5) = 2, p(5)=2p(5) = -2, m(5)=3m'(5) = 3, p(5)=1p'(5) = -1.
  • Substitute values: m(5)p(5)+m(5)p(5)=(2)(1)+(3)(2)=26=8m(5) p'(5) + m'(5) p(5) = (2)(-1) + (3)(-2) = -2 - 6 = -8 Answer: 8-8

(b) Find ddx[f(x)g(x)+h(x)p(x)]x=1\frac{d}{dx} [f(x) g(x) + h(x) p(x)] \Big|_{x=1}

Differentiate term by term: ddx[f(x)g(x)+h(x)p(x)]=f(x)g(x)+f(x)g(x)+h(x)p(x)+h(x)p(x)\frac{d}{dx} [f(x) g(x) + h(x) p(x)] = f'(x) g(x) + f(x) g'(x) + h'(x) p(x) + h(x) p'(x) Evaluate at x=1x = 1:

  • From the table, f(1)=3f(1) = 3, g(1)=2g(1) = 2, h(1)=1h(1) = -1, p(1)=2p(1) = -2, f(1)=1f'(1) = 1, g(1)=4g'(1) = 4, h(1)=3h'(1) = 3, p(1)=2p'(1) = 2.
  • Substitute values: f(1)g(1)+f(1)g(1)+h(1)p(1)+h(1)p(1)=(1)(2)+(3)(4)+(3)(2)+(1)(2)=2+1262=6f'(1) g(1) + f(1) g'(1) + h'(1) p(1) + h(1) p'(1) = (1)(2) + (3)(4) + (3)(-2) + (-1)(2) = 2 + 12 - 6 - 2 = 6 Answer: 66

(c) Find ddx[x2f(x)p(x)]x=1\frac{d}{dx} \left[ \frac{x^2 f(x)}{p(x)} \right] \Big|_{x=1}

Using the quotient rule: [ \frac{d}{dx} \left[ \frac{x^2 f(x)}{p(x)} \right] = \frac{(2x f(x) + x^2 f'(x)) p(x) - x^2 f(x) p'(x)}{[p(x)]^2} ] Evaluate at x=1x = 1:

  • From the table, f(1)=3f(1) = 3, p(1)=2p(1) = -2, f(1)=1f'(1) = 1, p(1)=2p'(1) = 2.
  • Substitute values: (213+121)(2)1232(2)2=(6+1)(2)324=144=3.5\frac{(2 \cdot 1 \cdot 3 + 1^2 \cdot 1)(-2) - 1^2 \cdot 3 \cdot 2}{(-2)^2} = \frac{(6 + 1)(-2) - 3 \cdot 2}{4} = \frac{-14}{4} = -3.5 Answer: 3.5-3.5

(d) Find ddx[xh(x)+4j(x)]x=4\frac{d}{dx} \left[ \sqrt{x \cdot h(x)} + 4 j(x) \right] \Big|_{x=4}

Using the chain rule and the product rule: ddx[xh(x)+4j(x)]=12xh(x)(h(x)+xh(x))+4j(x)\frac{d}{dx} \left[ \sqrt{x \cdot h(x)} + 4 j(x) \right] = \frac{1}{2 \sqrt{x h(x)}} (h(x) + x h'(x)) + 4 j'(x) Evaluate at x=4x = 4:

  • From the table, h(4)=3h(4) = -3, h(4)=2h'(4) = -2, j(x)=x3j(x)=3x2j(x) = x^3 \Rightarrow j'(x) = 3x^2, so j(4)=342=48j'(4) = 3 \cdot 4^2 = 48.
  • Substitute values: 124(3)(3+42)+448\frac{1}{2 \sqrt{4 \cdot (-3)}} ( -3 + 4 \cdot -2 ) + 4 \cdot 48 Note: Here, 4(3)\sqrt{4 \cdot (-3)} yields a complex value due to a negative radicand, which typically isn't allowed in a real-value calculus context. Verify this in class if needed.

Follow-Up Questions

  1. Would you like further details on any specific part?
  2. Do you want explanations for each derivative rule applied?
  3. Would you like to discuss the graph's role in solving any part?
  4. Should I proceed with the remaining questions in detail?
  5. Do you need help with interpreting the notation in these problems?

Tip: When using product and quotient rules, write down all values for each function and its derivative at a given xx-value to avoid errors in substitution.

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Math Problem Analysis

Mathematical Concepts

Derivatives
Product Rule
Quotient Rule
Chain Rule

Formulas

Product Rule: d/dx [u(x) v(x)] = u'(x)v(x) + u(x)v'(x)
Quotient Rule: d/dx [u(x)/v(x)] = [u'(x)v(x) - u(x)v'(x)] / [v(x)]^2
Chain Rule: d/dx [f(g(x))] = f'(g(x)) * g'(x)

Theorems

Product Rule Theorem
Quotient Rule Theorem
Chain Rule Theorem

Suitable Grade Level

Grades 11-12