Question

the derivative of $g(x)=e^{x}(13x^{2}-26x + 26)$ is
$g(x)=\frac{(13x^{2}-26x + 26)left(\frac{d}{dx}(e^{x})
ight)+(e^{x})left(\frac{d}{dx}(13x^{2}-26x + 26)
ight)}{(e^{x})^{2}}$

Explanation:

Step1: Apply product - rule

The product - rule states that if $g(x)=u(x)v(x)$, then $g^{\prime}(x)=u(x)v^{\prime}(x)+v(x)u^{\prime}(x)$. Here, let $u(x)=e^{x}$ and $v(x)=13x^{2}-26x + 26$.

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

The derivative of $u(x)=e^{x}$ with respect to $x$ is $u^{\prime}(x)=e^{x}$, since $\frac{d}{dx}(e^{x})=e^{x}$.

Step3: Find $v^{\prime}(x)$

Differentiate $v(x)=13x^{2}-26x + 26$ term - by - term. $\frac{d}{dx}(13x^{2})=26x$, $\frac{d}{dx}(-26x)=-26$, and $\frac{d}{dx}(26)=0$. So $v^{\prime}(x)=26x-26$.

Step4: Calculate $g^{\prime}(x)$

$g^{\prime}(x)=u(x)v^{\prime}(x)+v(x)u^{\prime}(x)=e^{x}(26x - 26)+(13x^{2}-26x + 26)e^{x}$.
Factor out $e^{x}$: $g^{\prime}(x)=e^{x}(26x-26 + 13x^{2}-26x + 26)=13x^{2}e^{x}$.

Answer:

$13x^{2}e^{x}$