Question

How do you find the derivative of `f(x) = (x^2 + 2x +3)e^-x`?

Answer 1

`d/dx((x^2+2x+3)e^-x)=-x^2e^-x-e^-x`

Explanation:

`d/dx((x^2+2x+3)e^-x)`
`=\underbrace(d/dx(x^2+2x+3))_((1)) * e^-x+(x^2+2x+3) * \underbrace(d/dx(e^-x))_((2)`
`\color(white)(...... ..............................)`(By the product rule)

Lets calculate `(1)` first:
`d/dx(x^2+2x+3)=d/dx(x^2)+d/dx(2x)+d/dx(3) = 2x+2+0=2x+2`

Now, let's calculate `(2)`:
`d/dx(e^-x)=d/(d(-x))e^-xd/dx(-x)`
`\color(white)(...... ..............................)`(By the chain rule)
`= e^-x*-1=-e^-x`

Combining the two, we get:
`(2x+2)* e^(-x)+(x^2+2x+3)*(-e^(-x))`
`=\cancel(2xe^-x)+2e^-x -x^2e^-x\cancel(-2xe^-x)-3e^-x`
`=-x^2e^-x-e^-x`