Math Problem Statement

y=2x/x^2+2 find dy/dx

Solution

The expression provided for yy is ambiguous because of how the division is written. It is unclear whether the entire expression is meant to be divided by x2+2x^2 + 2, or just the first term.

Option 1: If the expression is y=2xx2+2y = \frac{2x}{x^2 + 2}

In this case, we will apply the quotient rule to find the derivative dydx\frac{dy}{dx}.

The quotient rule states that if y=f(x)g(x)y = \frac{f(x)}{g(x)}, then: dydx=f(x)g(x)f(x)g(x)(g(x))2\frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} Here, f(x)=2xf(x) = 2x and g(x)=x2+2g(x) = x^2 + 2.

  1. f(x)=2f'(x) = 2
  2. g(x)=2xg'(x) = 2x

Now, applying the quotient rule: dydx=(2)(x2+2)(2x)(2x)(x2+2)2\frac{dy}{dx} = \frac{(2)(x^2 + 2) - (2x)(2x)}{(x^2 + 2)^2} Simplifying the numerator:

= \frac{2x^2 + 4 - 4x^2}{(x^2 + 2)^2} = \frac{-2x^2 + 4}{(x^2 + 2)^2}$$ So, the derivative is: $$\frac{dy}{dx} = \frac{-2x^2 + 4}{(x^2 + 2)^2}$$ ### Option 2: If the expression is $$y = \frac{2x}{x^2} + 2$$ This simplifies to: $$y = \frac{2}{x} + 2$$ Now, differentiate term by term: 1. The derivative of $$\frac{2}{x}$$ is $$-\frac{2}{x^2}$$. 2. The derivative of 2 is 0. Thus, the derivative is: $$\frac{dy}{dx} = -\frac{2}{x^2}$$ ### Clarification Needed Which interpretation is correct, Option 1 or Option 2? Let me know so I can confirm the solution! --- Would you like more details on any part of the solution? Here are 5 related questions to consider: 1. How is the quotient rule applied in calculus? 2. What is the difference between implicit and explicit differentiation? 3. How do you handle products and quotients in differentiation? 4. Can the chain rule be used here instead of the quotient rule? 5. What other methods could simplify rational functions before differentiation? **Tip:** Always clarify ambiguous expressions using parentheses, such as $$\frac{2x}{x^2 + 2}$$, to avoid confusion when differentiating.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Quotient Rule
Algebra

Formulas

Quotient rule: (f/g)' = (f'(x)g(x) - f(x)g'(x))/(g(x))^2
Term-by-term differentiation

Theorems

Quotient Rule
Power Rule

Suitable Grade Level

Grades 11-12 (Advanced High School) or College-level Calculus