If $f(x)=\frac{4-x^{2}}{7+x^{2}}$, find: \[ f^{\prime}(x)= \]

Step 1 ：Given the function \(f(x)=\frac{4-x^{2}}{7+x^{2}}\), we are asked to find its derivative \(f^{\prime}(x)\).

Step 2 ：We can use the quotient rule to find the derivative. The quotient rule states that the derivative of \(\frac{u}{v}\) is \(\frac{vu' - uv'}{v^2}\), where \(u\) and \(v\) are functions of \(x\), and \(u'\) and \(v'\) are their respective derivatives.

Step 3 ：In this case, \(u = 4 - x^2\) and \(v = 7 + x^2\).

Step 4 ：The derivative of \(u\) with respect to \(x\) is \(u' = -2x\) and the derivative of \(v\) with respect to \(x\) is \(v' = 2x\).

Step 5 ：Substituting these values into the quotient rule gives us \(f^{\prime}(x)=\frac{-2x(4 - x^{2}) - 2x(x^{2} + 7)}{(x^{2} + 7)^{2}}\).

Step 6 ：So, the derivative of the function \(f(x)=\frac{4-x^{2}}{7+x^{2}}\) is \(f^{\prime}(x)=\boxed{\frac{-2x(4 - x^{2}) - 2x(x^{2} + 7)}{(x^{2} + 7)^{2}}}\).