1. **State the problem:** We are given the function $$f(x) = \frac{2x}{x^2 + 1}$$ and need to find its first derivative $$f'(x)$$, second derivative $$f''(x)$$, and then evaluate these derivatives at $$x=1$$. 2. **Recall the formula:** For a function $$f(x) = \frac{u(x)}{v(x)}$$, the derivative is given by the quotient rule: $$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$ where $$u(x) = 2x$$ and $$v(x) = x^2 + 1$$. 3. **Find the first derivative $$f'(x)$$:** - Compute $$u'(x) = 2$$. - Compute $$v'(x) = 2x$$. - Apply the quotient rule: $$ f'(x) = \frac{2(x^2 + 1) - 2x(2x)}{(x^2 + 1)^2} = \frac{2x^2 + 2 - 4x^2}{(x^2 + 1)^2} = \frac{-2x^2 + 2}{(x^2 + 1)^2} = \frac{2(1 - x^2)}{(x^2 + 1)^2} $$ 4. **Find the second derivative $$f''(x)$$:** - Let $$g(x) = 2(1 - x^2)$$ and $$h(x) = (x^2 + 1)^2$$. - Then $$f'(x) = \frac{g(x)}{h(x)}$$. - Compute $$g'(x) = 2(-2x) = -4x$$. - Compute $$h'(x) = 2(x^2 + 1)(2x) = 4x(x^2 + 1)$$. - Apply the quotient rule again: $$ f''(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} = \frac{-4x (x^2 + 1)^2 - 2(1 - x^2) 4x (x^2 + 1)}{(x^2 + 1)^4} $$ - Simplify numerator: $$ -4x (x^2 + 1)^2 - 8x (1 - x^2)(x^2 + 1) = -4x (x^2 + 1) [ (x^2 + 1) + 2(1 - x^2) ] $$ - Inside brackets: $$ (x^2 + 1) + 2(1 - x^2) = x^2 + 1 + 2 - 2x^2 = 3 - x^2 $$ - So numerator is: $$ -4x (x^2 + 1)(3 - x^2) $$ - Therefore: $$ f''(x) = \frac{-4x (x^2 + 1)(3 - x^2)}{(x^2 + 1)^4} = \frac{-4x (3 - x^2)}{(x^2 + 1)^3} $$ 5. **Evaluate $$f'(1)$$:** $$ f'(1) = \frac{2(1 - 1^2)}{(1^2 + 1)^2} = \frac{2(1 - 1)}{(1 + 1)^2} = \frac{0}{4} = 0 $$ 6. **Evaluate $$f''(1)$$:** $$ f''(1) = \frac{-4(1)(3 - 1^2)}{(1^2 + 1)^3} = \frac{-4(1)(3 - 1)}{(1 + 1)^3} = \frac{-4 \times 2}{8} = -1 $$ **Final answers:** $$ f'(x) = \frac{2(1 - x^2)}{(x^2 + 1)^2}, \quad f''(x) = \frac{-4x(3 - x^2)}{(x^2 + 1)^3}, \quad f'(1) = 0, \quad f''(1) = -1 $$