Math Problem Statement

Given the graphs of f(x) and g(x), find the following derivatives using the chain rule and quotient rule.

Solution

The image you uploaded shows a problem that involves applying the chain rule for derivatives. It includes the graphs of two functions f(x)f(x) and g(x)g(x), and a few derivative problems based on these graphs. Let me break it down and provide solutions.

Problem Breakdown:

  • Given Functions:

    • g(x)=2g(x) = 2, f(a)=3f(a) = -3 are some values provided.
    • You have graphs of f(x)f(x) and g(x)g(x) on the left side of the sheet, though their specific behavior needs to be analyzed visually to extract function values and slopes.
  • Tasks:

    1. Chain Rule: ddx[g(f(x))]=g(f(x))f(x)\frac{d}{dx}[g(f(x))] = g'(f(x)) f'(x) This is the formula needed for parts (a) and (b) under the chain rule, where you must apply the derivative to the composite function.

    2. Parts (a) and (b): Find ddx[g(f(x))]\frac{d}{dx}[g(f(x))] for specific values of xx (1 and -1).

    3. Part (d): Show ddx[g(x)f2(x)]\frac{d}{dx} \left[ \frac{\sqrt{g(x)}}{f_2(x)} \right] at x=5x = 5.


Part (a):

Find ddx[g(f(x))]\frac{d}{dx}[g(f(x))] at x=1x = 1.

By applying the chain rule: ddx[g(f(x))]=g(f(x))f(x)\frac{d}{dx}[g(f(x))] = g'(f(x)) f'(x)

  • Evaluate f(1)f(1) from the graph.
  • Find f(1)f'(1), the slope of the tangent to the graph of f(x)f(x) at x=1x = 1.
  • Use g(f(1))g'(f(1)), which depends on the value of f(1)f(1).

Part (b):

Find ddx[g(f(x))]\frac{d}{dx}[g(f(x))] at x=1x = -1.

Similarly, apply the chain rule: ddx[g(f(x))]=g(f(x))f(x)\frac{d}{dx}[g(f(x))] = g'(f(x)) f'(x)

  • Evaluate f(1)f(-1) and find f(1)f'(-1) using the graph.
  • Use g(f(1))g'(f(-1)) to complete the calculation.

Part (d):

Show that ddx[g(x)f2(x)]=28\frac{d}{dx} \left[ \frac{\sqrt{g(x)}}{f_2(x)} \right] = \frac{\sqrt{2}}{8}.

For this, apply the quotient rule combined with the chain rule.

  • Use the formula for the derivative of a quotient: ddx[u(x)v(x)]=v(x)u(x)u(x)v(x)v(x)2\frac{d}{dx} \left[ \frac{u(x)}{v(x)} \right] = \frac{v(x)u'(x) - u(x)v'(x)}{v(x)^2}
  • Set u(x)=g(x)u(x) = \sqrt{g(x)} and v(x)=f2(x)v(x) = f_2(x).
  • Find the derivatives u(x)u'(x) and v(x)v'(x), and plug them into the quotient rule formula.

Do you want me to walk through the solutions to specific parts of the problem in more detail?


Follow-up Questions:

  1. How do we apply the chain rule to composite functions involving graphs?
  2. How do we find f(x)f'(x) and g(x)g'(x) from a graph?
  3. What is the difference between applying the chain rule and the quotient rule?
  4. How can we interpret a derivative in terms of rates of change using graphical information?
  5. What are the steps for solving a derivative problem with multiple rules (e.g., chain + quotient)?

Tip: When solving derivatives with both chain and quotient rules, always start by simplifying the expressions as much as possible before applying the rules.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Chain Rule
Quotient Rule

Formulas

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

Theorems

Chain Rule
Quotient Rule

Suitable Grade Level

Grade 11-12 (High School Calculus)