Question

the derivative of $g(x)=e^{x}(13x^{2}-26x + 20)$ is $\square$

Explanation:

Step1: Apply product - rule

The product - rule states that if \(y = u\cdot v\), then \(y^\prime=u^\prime v + uv^\prime\). Let \(u = e^{x}\) and \(v = 13x^{2}-26x + 20\). The derivative of \(u = e^{x}\) is \(u^\prime=e^{x}\), and the derivative of \(v = 13x^{2}-26x + 20\) is \(v^\prime=26x-26\).

Step2: Calculate \(g^\prime(x)\)

\[

$$\begin{align*} g^\prime(x)&=u^\prime v+uv^\prime\\ &=e^{x}(13x^{2}-26x + 20)+e^{x}(26x - 26)\\ &=e^{x}(13x^{2}-26x + 20+26x - 26)\\ &=e^{x}(13x^{2}-6) \end{align*}$$

\]

Answer:

\(e^{x}(13x^{2}-6)\)