nd \( f^{\prime}(x), f^{\prime \prime}(x), f^{\prime}(1) \) and \( f^{\prime \prime}(1) \) when \( f(x)=\frac{2 x}{x^{2}+1} \)

First, let's find \( f^{\prime}(x) \). Using the quotient rule, the derivative of \( f(x) = \frac{2x}{x^2 + 1} \) is calculated as follows: \[ f^{\prime}(x) = \frac{(2)(x^2 + 1) - (2x)(2x)}{(x^2 + 1)^2} = \frac{2(x^2 + 1) - 4x^2}{(x^2 + 1)^2} = \frac{2 - 2x^2}{(x^2 + 1)^2} = \frac{2(1 - x^2)}{(x^2 + 1)^2} \] Next, we calculate \( f^{\prime}(1) \): \[ f^{\prime}(1) = \frac{2(1 - 1^2)}{(1^2 + 1)^2} = \frac{2(1 - 1)}{(1 + 1)^2} = \frac{0}{4} = 0 \] Now, let's find \( f^{\prime \prime}(x) \). Differentiate \( f^{\prime}(x) \) using the quotient rule again: \[ f^{\prime \prime}(x) = \frac{(4x)(x^2 + 1)^2 - 2(1 - x^2)(2(x^2 + 1)(2x))}{(x^2 + 1)^4} \] Now, simplifying for \( f^{\prime \prime}(1) \): First, evaluate \( f^{\prime \prime}(x) \) at \( x = 1 \): 1. Evaluate \( (4 \cdot 1)(1^2 + 1)^2 = 4 \cdot 4 = 16 \) 2. For the second part, we need \( 1 - 1^2 = 0 \), so the entire second part is 0. Combining these results gives: \[ f^{\prime \prime}(1) = \frac{16 - 0}{(2)^4} = \frac{16}{16} = 1 \] Thus, we conclude with: \[ f^{\prime}(1) = 0 \quad \text{and} \quad f^{\prime \prime}(1) = 1 \] To summarize: - \( f^{\prime}(x) = \frac{2(1 - x^2)}{(x^2 + 1)^2} \) - \( f^{\prime}(1) = 0 \) - \( f^{\prime \prime}(1) = 1 \)