Find the derivative of the given function. \[ f(x)=\frac{x+1}{x^{2}+1} \] \( f^{\prime}(x)=\frac{[?] x^{2}+\square x+\square}{\left(x^{2}+1\right)^{2}} \)

To find the derivative of the function \( f(x)=\frac{x+1}{x^{2}+1} \), we can use the quotient rule, which states that if we have a function in the form \( \frac{u}{v} \), the derivative \( f'(x) \) is given by \( \frac{u'v - uv'}{v^2} \). Here, \( u = x + 1 \) and \( v = x^2 + 1 \). Calculating the derivatives: - \( u' = 1 \) (the derivative of \( x + 1 \)) - \( v' = 2x \) (the derivative of \( x^2 + 1 \)) Now applying the quotient rule: \[ f'(x) = \frac{(1)(x^2 + 1) - (x + 1)(2x)}{(x^2 + 1)^2} \] Expanding the numerator: \[ f'(x) = \frac{x^2 + 1 - (2x^2 + 2x)}{(x^2 + 1)^2} = \frac{x^2 + 1 - 2x^2 - 2x}{(x^2 + 1)^2} = \frac{-x^2 - 2x + 1}{(x^2 + 1)^2} \] Thus, we can express the derivative as: \[ f'(x) = \frac{-x^2 - 2x + 1}{(x^2 + 1)^2} \] Therefore, according to your format, we fill in the blanks as follows: \[ f^{\prime}(x)=\frac{[-1] x^{2}+\square x+\square}{\left(x^{2}+1\right)^{2}} \] Where the second blank is \(-2\) for \( -2x \) and the last blank is \(1\) for the constant term. So the filled form is: \[ f^{\prime}(x)=\frac{[-1] x^{2}+\mathbf{-2} x+\mathbf{1}}{\left(x^{2}+1\right)^{2}} \]