Answer:
the question is incomplete, the complete question is "Finding second Derivatives In Exercise,find the second derivate.
[tex]f(x)=(3+2x)e^{-3x}[/tex]"
answer:[tex] \frac{df(x)^{2}}{dx^{2}}=(15+18x)e^{-3x}\\[/tex],
Step-by-step explanation:
To determine the second derivative, we differentiate twice.
for the first differentiation, we use the product rule approach. i.e
[tex]f(x)=u(x)v(x)\\\frac{df(x)}{dx}=u(x)\frac{dv(x)}{dx}+ v(x)\frac{du(x)}{dx}\\[/tex]
from [tex]f(x)=(3+2x)e^{-3x}[/tex] if w assign
u(x)=(3+2x) and the derivative, [tex] \frac{du(x)}{dx}=2[/tex]
also [tex]v(x)=e^{-3x}[/tex] and the derivative [tex] \frac{dv(x)}{dx}=-3e^{-3x}[/tex].
If we substitute values we arrive at
[tex]\frac{df(x)}{dx}=(3+2x)(-3e^{-3x})+2e^{-3x}\\\frac{df(x)}{dx}=(-9-6x)e^{-3x}+2e^{-3x}\\\frac{df(x)}{dx}=(-9-6x+2)e^{-3x}\\\frac{df(x)}{dx}=-(7+6x)e^{-3x}\\[/tex],
Now to determine the second derivative we use the product rule again
this time, u(x)=(7+6x) and the derivative, [tex] \frac{du(x)}{dx}=6[/tex]
also [tex]v(x)=e^{-3x}[/tex] and the derivative [tex] \frac{dv(x)}{dx}=-3e^{-3x}[/tex].
If we substitute values we arrive at
[tex]\frac{df(x)^{2}}{dx^{2}}=-((7+6x)(-3e^{-3x})+6e^{-3x})\\\frac{df(x)^{2}}{dx^{2}}=-((-21-18x)e^{-3x}+6e^{-3x})\\\frac{df(x)^{2}}{dx^{2}}=-((-21-18x+6)e^{-3x})\\\frac{df(x)^{2}}{dx^{2}}=(15+18x)e^{-3x}\\[/tex],
