Question

hw11 differentiation rules ii (target c1, c2, c5; §3.3)
score: 3/8 answered: 3/8
question 4
find the derivative of the function $g(x)=\frac{e^{x}}{4 - 4x}$
$g(x)=$
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Explanation:

Step1: Recall quotient - rule

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

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 $v(x)=4 - 4x$ is $v^{\prime}(x)=-4$.

Step3: Apply the quotient - rule

\[

$$\begin{align*} g^{\prime}(x)&=\frac{e^{x}(4 - 4x)-e^{x}(-4)}{(4 - 4x)^{2}}\\ &=\frac{e^{x}(4 - 4x + 4)}{(4 - 4x)^{2}}\\ &=\frac{e^{x}(8 - 4x)}{(4 - 4x)^{2}}\\ &=\frac{4e^{x}(2 - x)}{16(1 - x)^{2}}\\ &=\frac{e^{x}(2 - x)}{4(1 - x)^{2}} \end{align*}$$

\]

Answer:

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