Question

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find the derivative of the function.
f(x) = (x + 1)/(4x^2 + 6x + 7)
f(x) =
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Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $f(x)=\frac{u(x)}{v(x)}$, then $f^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v(x)^2}$. Here, $u(x)=x + 1$, so $u^{\prime}(x)=1$, and $v(x)=4x^{2}+6x + 7$, so $v^{\prime}(x)=8x+6$.

Step2: Substitute into quotient - rule

\[

$$\begin{align*} f^{\prime}(x)&=\frac{1\times(4x^{2}+6x + 7)-(x + 1)\times(8x+6)}{(4x^{2}+6x + 7)^{2}}\\ &=\frac{4x^{2}+6x + 7-(8x^{2}+6x+8x + 6)}{(4x^{2}+6x + 7)^{2}}\\ &=\frac{4x^{2}+6x + 7-(8x^{2}+14x + 6)}{(4x^{2}+6x + 7)^{2}}\\ &=\frac{4x^{2}+6x + 7 - 8x^{2}-14x - 6}{(4x^{2}+6x + 7)^{2}}\\ &=\frac{-4x^{2}-8x + 1}{(4x^{2}+6x + 7)^{2}} \end{align*}$$

\]

Answer:

$\frac{-4x^{2}-8x + 1}{(4x^{2}+6x + 7)^{2}}$