Question

Finding second Derivatives In Exercise,find the second derivate. f(x) = (3 + 2x)e - 3x

Answer 1

Answer:

the question is incomplete, the complete question is "Finding second Derivatives In Exercise,find the second derivate.

$f(x)=(3+2x)e^{-3x}$"

answer:$\frac{df(x)^{2}}{dx^{2}}=(15+18x)e^{-3x}$,

Step-by-step explanation:

To determine the second derivative, we differentiate twice.

for the first differentiation, we use the product rule approach. i.e

$f(x)=u(x)v(x)\\\frac{df(x)}{dx}=u(x)\frac{dv(x)}{dx}+ v(x)\frac{du(x)}{dx}$

from $f(x)=(3+2x)e^{-3x}$ if w assign

u(x)=(3+2x)  and the derivative, $\frac{du(x)}{dx}=2$

also $v(x)=e^{-3x}$ and the derivative  $\frac{dv(x)}{dx}=-3e^{-3x}$.

If we substitute values we arrive at

$\frac{df(x)}{dx}=(3+2x)(-3e^{-3x})+2e^{-3x}\\\frac{df(x)}{dx}=(-9-6x)e^{-3x}+2e^{-3x}\\\frac{df(x)}{dx}=(-9-6x+2)e^{-3x}\\\frac{df(x)}{dx}=-(7+6x)e^{-3x}$,

Now to determine the second derivative we use the product rule again

this time, u(x)=(7+6x)  and the derivative, $\frac{du(x)}{dx}=6$

also $v(x)=e^{-3x}$ and the derivative  $\frac{dv(x)}{dx}=-3e^{-3x}$.

If we substitute values we arrive at

$\frac{df(x)^{2}}{dx^{2}}=-((7+6x)(-3e^{-3x})+6e^{-3x})\\\frac{df(x)^{2}}{dx^{2}}=-((-21-18x)e^{-3x}+6e^{-3x})\\\frac{df(x)^{2}}{dx^{2}}=-((-21-18x+6)e^{-3x})\\\frac{df(x)^{2}}{dx^{2}}=(15+18x)e^{-3x}$,