Math Problem Statement

y = (x)/(x^2 +4) find the derivative

Solution

To find the derivative of the function

y=xx2+4y = \frac{x}{x^2 + 4}

we will use the quotient rule, which states that if we have a function in the form y=f(x)g(x)y = \frac{f(x)}{g(x)}, then the derivative yy' is given by:

y=f(x)g(x)f(x)g(x)[g(x)]2y' = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}

Here:

  • f(x)=xf(x) = x and g(x)=x2+4g(x) = x^2 + 4
  • f(x)=1f'(x) = 1
  • g(x)=2xg'(x) = 2x

Let's apply the quotient rule:

y=(1)(x2+4)(x)(2x)(x2+4)2y' = \frac{(1)(x^2 + 4) - (x)(2x)}{(x^2 + 4)^2}

Now, simplify the expression:

  1. Expand the terms in the numerator:

    y=x2+42x2(x2+4)2y' = \frac{x^2 + 4 - 2x^2}{(x^2 + 4)^2}

  2. Combine like terms:

    y=x2+4(x2+4)2=4x2(x2+4)2y' = \frac{-x^2 + 4}{(x^2 + 4)^2} = \frac{4 - x^2}{(x^2 + 4)^2}

So, the derivative of y=xx2+4y = \frac{x}{x^2 + 4} is:

y=4x2(x2+4)2y' = \frac{4 - x^2}{(x^2 + 4)^2}

Would you like additional details or have any questions?


Here are 5 related questions:

  1. How can we find the second derivative of y=xx2+4y = \frac{x}{x^2 + 4}?
  2. What is the derivative of y=xx2+cy = \frac{x}{x^2 + c} for a general constant cc?
  3. What are the critical points of y=xx2+4y = \frac{x}{x^2 + 4}?
  4. How does the derivative change if we add a constant to the numerator?
  5. Can we use logarithmic differentiation on this function? If so, how?

Tip: The quotient rule can be simplified by rewriting the function if possible, but it's usually best for functions with more complex denominators.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Quotient Rule

Formulas

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

Theorems

Quotient Rule

Suitable Grade Level

Grades 11-12