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

Question

Paisley Johnson · Accepted Answer

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

Cameron Alexander · Answer

undefined To find the derivative of $ f(x) = (x^2 + 2x +3)e^{-x} $, you can use the product rule and the chain rule. The product rule states that if $ f(x) = u(x)v(x) $, then $ f'(x) = u'(x)v(x) + u(x)v'(x) $, where $ u(x) $ and $ v(x) $ are differentiable functions of $ x $. The chain rule states that if $ y = f(g(x)) $, then $ \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} $, where $ y $ is a function of $ u $, and $ u $ is a function of $ x $.

Using the product rule and the chain rule, the derivative of $ f(x) $ is:

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

Simplifying this expression gives:

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

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