find the derivative of the function $g(x)=\frac{e^{x}}{4 + 3x}$\n$g(x)=$

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $g(x)=\frac{u(x)}{v(x)}$, then $g^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v(x)^2}$. Here, $u(x)=e^{x}$ and $v(x)=4 + 3x$.

Step2: Find $u^{\prime}(x)$ and $v^{\prime}(x)$

The derivative of $u(x)=e^{x}$ is $u^{\prime}(x)=e^{x}$, and the derivative of $v(x)=4 + 3x$ is $v^{\prime}(x)=3$.

Step3: Apply the quotient - rule

Substitute $u(x),u^{\prime}(x),v(x),v^{\prime}(x)$ into the quotient - rule formula: [ \begin{align*} g^{\prime}(x)&=\frac{e^{x}(4 + 3x)-e^{x}\times3}{(4 + 3x)^2}\ &=\frac{e^{x}(4 + 3x-3)}{(4 + 3x)^2}\ &=\frac{e^{x}(3x + 1)}{(4 + 3x)^2} \end{align*} ]

Answer:

$\frac{e^{x}(3x + 1)}{(4 + 3x)^2}$