Calculate $\frac{d y}{d x}$. You need not expand your answer. \[ \begin{array}{l} y=\frac{7 x^{2}-9 x+13}{2 x+4} \\ \frac{d y}{d x}=\square \end{array} \]
Solution
Step 1 ：Given the function \(y=\frac{7 x^{2}-9 x+13}{2 x+4}\), we are asked to find the derivative of this function.
Step 2 ：We will use the quotient rule for differentiation, which 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, \(u = 7x^2 - 9x + 13\) and \(v = 2x + 4\).
Step 4 ：First, we find the derivatives of \(u\) and \(v\). The derivative of \(u\) is \(u' = 14x - 9\) and the derivative of \(v\) is \(v' = 2\).
Step 5 ：Now that we have \(u'\) and \(v'\), we can substitute these into the quotient rule formula to find \(\frac{dy}{dx}\).
Step 6 ：Substituting the values into the formula, we get \(\frac{dy}{dx} = \frac{-14x^{2} + 18x + (2x + 4)(14x - 9) - 26}{(2x + 4)^{2}}\).
Step 7 ：\(\boxed{\frac{dy}{dx} = \frac{-14x^{2} + 18x + (2x + 4)(14x - 9) - 26}{(2x + 4)^{2}}}\) is the derivative of the given function.
