Question

find the derivative of the following function. g(x)=e^x(3x^2 + 2x + 7). g(x)=

Explanation:

Step1: Apply product - rule

The product - rule states that if $g(x)=u(x)v(x)$, then $g'(x)=u'(x)v(x)+u(x)v'(x)$. Here, $u(x)=e^{x}$ and $v(x)=3x^{2}+2x + 7$.

Step2: Find $u'(x)$

The derivative of $u(x)=e^{x}$ is $u'(x)=e^{x}$.

Step3: Find $v'(x)$

Differentiate $v(x)=3x^{2}+2x + 7$ term - by - term. The derivative of $3x^{2}$ is $6x$ (using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$), the derivative of $2x$ is $2$, and the derivative of the constant $7$ is $0$. So $v'(x)=6x + 2$.

Step4: Calculate $g'(x)$

Substitute $u(x),u'(x),v(x),v'(x)$ into the product - rule formula:
$g'(x)=e^{x}(3x^{2}+2x + 7)+e^{x}(6x + 2)$.
Factor out $e^{x}$: $g'(x)=e^{x}(3x^{2}+2x + 7+6x + 2)=e^{x}(3x^{2}+8x + 9)$.

Answer:

$e^{x}(3x^{2}+8x + 9)$