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Mathematics

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Answer

4x(x2+1)2\frac{4x}{(x^2 + 1)^2}

Step-by-step solution

6 steps

To find the derivative of f(x)=x2−1x2+1f(x) = \frac{x^2 - 1}{x^2 + 1}, we will use the quotient rule. The quotient rule states that if f(x)=g(x)h(x)f(x) = \frac{g(x)}{h(x)}, then f′(x)=g′(x)h(x)−g(x)h′(x)(h(x))2f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}.

  1. Identify g(x)g(x) and h(x)h(x).

    Let g(x)=x2−1g(x) = x^2 - 1.
    Let h(x)=x2+1h(x) = x^2 + 1.

  2. Find the derivative of g(x)g(x), denoted as g′(x)g'(x).

    g′(x)=ddx(x2−1)=2xg'(x) = \frac{d}{dx}(x^2 - 1) = 2x

  3. Find the derivative of h(x)h(x), denoted as h′(x)h'(x).

    h′(x)=ddx(x2+1)=2xh'(x) = \frac{d}{dx}(x^2 + 1) = 2x

  4. Apply the quotient rule formula.

    f′(x)=g′(x)h(x)−g(x)h′(x)(h(x))2f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}

    Substitute the expressions for g(x)g(x), h(x)h(x), g′(x)g'(x), and h′(x)h'(x):

    f′(x)=(2x)(x2+1)−(x2−1)(2x)(x2+1)2f'(x) = \frac{(2x)(x^2 + 1) - (x^2 - 1)(2x)}{(x^2 + 1)^2}

  5. Expand and simplify the numerator.

    Numerator=2x(x2+1)−(x2−1)(2x)\text{Numerator} = 2x(x^2 + 1) - (x^2 - 1)(2x) Numerator=(2x3+2x)−(2x3−2x)\text{Numerator} = (2x^3 + 2x) - (2x^3 - 2x) Numerator=2x3+2x−2x3+2x\text{Numerator} = 2x^3 + 2x - 2x^3 + 2x Numerator=4x\text{Numerator} = 4x

  6. Write the final derivative.

    f′(x)=4x(x2+1)2f'(x) = \frac{4x}{(x^2 + 1)^2}

    The derivative of f(x)=x2−1x2+1f(x) = \frac{x^2 - 1}{x^2 + 1} is 4x(x2+1)2\boxed{\frac{4x}{(x^2 + 1)^2}}.

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