13. find the derivative of f(x)=axe^(-bx + 7). assume that a and b are constants. f(x)=

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = uv$, where $u$ and $v$ are functions of $x$, then $y'=u'v + uv'$. Let $u = ax$ and $v = e^{-bx + 7}$. First, find $u'$ and $v'$. The derivative of $u = ax$ with respect to $x$ is $u'=a$ (since the derivative of $kx$ with respect to $x$ is $k$ for a constant $k$).

Step2: Find derivative of $v$

To find the derivative of $v = e^{-bx + 7}$, use the chain - rule. If $y = e^{u}$ and $u=-bx + 7$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. The derivative of $e^{u}$ with respect to $u$ is $e^{u}$, and the derivative of $u=-bx + 7$ with respect to $x$ is $-b$. So, $v'=-be^{-bx + 7}$.

Step3: Calculate $f'(x)$

By the product - rule $f'(x)=u'v+uv'$. Substitute $u = ax$, $u' = a$, $v = e^{-bx + 7}$, and $v'=-be^{-bx + 7}$ into the product - rule formula: [ \begin{align*} f'(x)&=a\cdot e^{-bx + 7}+ax\cdot(-b)e^{-bx + 7}\ &=ae^{-bx + 7}-abxe^{-bx + 7}\ &=ae^{-bx + 7}(1 - bx) \end{align*} ]

Answer:

$ae^{-bx + 7}(1 - bx)$