Math Problem Statement

Diketahui f(x) = x + 1 dan g(x) = x^2 - 1. Rumus fungsi (f/g)'(x) adalah ....

Solution

The question asks for the formula for the derivative of the division of two functions, f(x)f(x) and g(x)g(x), where:

f(x)=x+1andg(x)=x21.f(x) = x + 1 \quad \text{and} \quad g(x) = x^2 - 1.

To find fg(x)\frac{f}{g}'(x), we use the quotient rule for differentiation:

(fg)(x)=f(x)g(x)f(x)g(x)[g(x)]2.\left( \frac{f}{g} \right)'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}.

Steps:

  1. Find f(x)f'(x): f(x)=ddx(x+1)=1.f'(x) = \frac{d}{dx}(x + 1) = 1.

  2. Find g(x)g'(x): g(x)=ddx(x21)=2x.g'(x) = \frac{d}{dx}(x^2 - 1) = 2x.

  3. Substitute f(x),f(x),g(x),g(x)f(x), f'(x), g(x), g'(x) into the quotient rule: (fg)(x)=(1)(x21)(x+1)(2x)(x21)2.\left( \frac{f}{g} \right)'(x) = \frac{(1)(x^2 - 1) - (x + 1)(2x)}{(x^2 - 1)^2}.

  4. Simplify the numerator: (1)(x21)(x+1)(2x)=x212x22x=x22x1.(1)(x^2 - 1) - (x + 1)(2x) = x^2 - 1 - 2x^2 - 2x = -x^2 - 2x - 1.

  5. Final formula: (fg)(x)=x22x1(x21)2.\left( \frac{f}{g} \right)'(x) = \frac{-x^2 - 2x - 1}{(x^2 - 1)^2}.

The derivative of f(x)g(x)\frac{f(x)}{g(x)} is:

x22x1(x21)2\boxed{\frac{-x^2 - 2x - 1}{(x^2 - 1)^2}}

Would you like a more detailed explanation of any step? Here are five related questions to consider:

  1. How is the quotient rule applied to general functions f(x)f(x) and g(x)g(x)?
  2. Why is it necessary to calculate both f(x)f'(x) and g(x)g'(x)?
  3. How do we simplify the numerator in the quotient rule?
  4. Can this formula be further factorized or simplified?
  5. What happens if g(x)=0g(x) = 0 at certain points?

Tip: Always double-check your simplifications, especially in the numerator, to ensure the final formula is correct!

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Math Problem Analysis

Mathematical Concepts

Quotient Rule
Differentiation

Formulas

Quotient Rule: (f/g)'(x) = [f'(x)g(x) - f(x)g'(x)] / [g(x)]^2

Theorems

Differentiation rules

Suitable Grade Level

Grades 10-12

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