\( \begin{array}{l} \text { Find the derivative of the function. }\\ y=\frac{7 x}{\sqrt{x^{2}+4}} \end{array} \)

To find the derivative of the function \( y=\frac{7 x}{\sqrt{x^{2}+4}} \), we will use the quotient rule. The quotient rule states that if you have a function \( y = \frac{u}{v} \), then \( y' = \frac{u'v - uv'}{v^2} \). Here, let \( u = 7x \) and \( v = \sqrt{x^2 + 4} \). Calculating the derivatives: - \( u' = 7 \) - To find \( v' \), we apply the chain rule: \[ v = (x^2 + 4)^{1/2} \Rightarrow v' = \frac{1}{2}(x^2 + 4)^{-1/2}(2x) = \frac{x}{\sqrt{x^2 + 4}} \] Now substituting back into the quotient rule: \[ y' = \frac{(7)(\sqrt{x^2 + 4}) - (7x)\left(\frac{x}{\sqrt{x^2 + 4}}\right)}{(x^2 + 4)} \] Simplifying this gives: \[ y' = \frac{7\sqrt{x^2 + 4} - \frac{7x^2}{\sqrt{x^2 + 4}}}{x^2 + 4} \] Combining terms in the numerator leads to: \[ y' = \frac{7(x^2 + 4) - 7x^2}{(x^2 + 4)\sqrt{x^2 + 4}} = \frac{28}{(x^2 + 4)\sqrt{x^2 + 4}} \] So the derivative is: \[ y' = \frac{28}{(x^2 + 4)\sqrt{x^2 + 4}} \]