Math Problem Statement
Solution
Let's solve the three parts of the problem step by step using the provided information about , , , and at .
The table gives:
x &= 2, \, f(2) = 8, \, g(2) = 2, \, f'(2) = 2\pi, \, g'(2) = e. \end{aligned}$$ --- ### Part (a): $$\frac{d}{dx} \left[ f(g(x)) \right]$$ at $$x = 2$$ To differentiate $$f(g(x))$$, we use the **chain rule**: $$\frac{d}{dx} \left[ f(g(x)) \right] = f'(g(x)) \cdot g'(x).$$ At $$x = 2$$: - $$g(2) = 2$$ (from the table), so $$f'(g(2)) = f'(2) = 2\pi$$. - $$g'(2) = e$$. Thus, $$\frac{d}{dx} \left[ f(g(x)) \right] \Big|_{x=2} = f'(g(2)) \cdot g'(2) = (2\pi) \cdot e = 2\pi e.$$ --- ### Part (b): $$\frac{d}{dx} \left[ \frac{1}{f(x)} \right]$$ at $$x = 2$$ To differentiate $$\frac{1}{f(x)}$$, use the **quotient rule** or the derivative of $$x^{-1}$$: \[ \frac{d}{dx} \left[ \frac{1}{f(x)} \right] = -\frac{f'(x)}{[f(x)]^2}. \] At $$x = 2$$: - $$f(2) = 8$$, - $$f'(2) = 2\pi$$. Thus, \[ \frac{d}{dx} \left[ \frac{1}{f(x)} \right] \Big|_{x=2} = -\frac{f'(2)}{[f(2)]^2} = -\frac{2\pi}{8^2}. \] Simplify: $$\frac{d}{dx} \left[ \frac{1}{f(x)} \right] \Big|_{x=2} = -\frac{2\pi}{64} = -\frac{\pi}{32}.$$ --- ### Part (c): $$\frac{d}{dx} \left[ \frac{g(x)}{f(x)} \right]$$ at $$x = 2$$ To differentiate $$\frac{g(x)}{f(x)}$$, use the **quotient rule**: \[ \frac{d}{dx} \left[ \frac{g(x)}{f(x)} \right] = \frac{g'(x)f(x) - g(x)f'(x)}{[f(x)]^2}. \] At $$x = 2$$: - $$g'(2) = e$$, - $$f(2) = 8$$, - $$g(2) = 2$$, - $$f'(2) = 2\pi$$. Substitute these values: $$\frac{d}{dx} \left[ \frac{g(x)}{f(x)} \right] \Big|_{x=2} = \frac{(e)(8) - (2)(2\pi)}{8^2}.$$ Simplify the numerator: $$(e)(8) - (2)(2\pi) = 8e - 4\pi.$$ The denominator is: $$8^2 = 64.$$ Thus, $$\frac{d}{dx} \left[ \frac{g(x)}{f(x)} \right] \Big|_{x=2} = \frac{8e - 4\pi}{64}.$$ Simplify further: $$\frac{d}{dx} \left[ \frac{g(x)}{f(x)} \right] \Big|_{x=2} = \frac{8e}{64} - \frac{4\pi}{64} = \frac{e}{8} - \frac{\pi}{16}.$$ --- ### Final Answers: 1. **(a)** $$\frac{d}{dx} \left[ f(g(x)) \right] \Big|_{x=2} = 2\pi e$$. 2. **(b)** $$\frac{d}{dx} \left[ \frac{1}{f(x)} \right] \Big|_{x=2} = -\frac{\pi}{32}$$. 3. **(c)** $$\frac{d}{dx} \left[ \frac{g(x)}{f(x)} \right] \Big|_{x=2} = \frac{e}{8} - \frac{\pi}{16}$$. --- ### Do you want further clarifications or step-by-step explanations for any part? ### Related Questions: 1. How do you apply the chain rule for composite functions in general? 2. What is the quotient rule, and how is it derived? 3. How do you find critical points of functions involving $$\frac{g(x)}{f(x)}$$? 4. What are higher-order derivatives of composite functions like $$f(g(x))$$? 5. How can the derivative rules be extended for functions like $$\frac{1}{f(x)}$$ or $$g(x)^n$$? **Tip:** Always check whether the functions $$f(x)$$ or $$g(x)$$ are defined and differentiable at the given point before applying derivative rules.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Differentiation
Chain Rule
Quotient Rule
Derivatives of Composite Functions
Formulas
Chain Rule: d/dx [f(g(x))] = f'(g(x)) * g'(x)
Derivative of 1/f(x): d/dx [1/f(x)] = -f'(x) / [f(x)]^2
Quotient Rule: d/dx [g(x)/f(x)] = [g'(x)f(x) - g(x)f'(x)] / [f(x)]^2
Theorems
Chain Rule
Quotient Rule
Suitable Grade Level
Advanced High School Calculus or Early University Level
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