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

Question

Charles Anctil · Accepted Answer

#dy/dx=xe^-x(2-x).# #y=x^2e^-x# #:. dy/dx=x^2*d/dx(e^-x)+e^-x*d/dxx^2#.... [Product Rule for Diffn.] #=x^2*e^-x*d/dx(-x)+e^-x*2x#...................[Chain rule] #=-x^2e^-x+2x*e^-x=xe^-x(2-x).#

William Abdalla · Answer

#xe^(-x)(2-x)# Differentiate using the #color(blue)"product rule"# #color(red)(|bar(ul(color(white)(a/a)color(black)(f(x)=g(x)h(x)rArrf'(x)=g(x)h'(x)+h(x)g'(x))color(white)(a/a)|)))# here #g(x)=x^2rArrg'(x)=2x# and #h(x)=e^(-x)rArrh'(x)=e^(-x) (-1)=-e^(-x)# #"-------------------------------------------------------------------"# Substitute these values into f'(x) #f'(x)=x^2(-e^(-x))+e^(-x)(2x)=-x^2e^(-x)+2xe^(-x)# #rArrdy/dx=xe^(-x)(2-x)#

Adam Adkins · Answer

undefined To find the derivative of $ y = x^2 e^{-x} $, you can use the product rule. The product rule states that if $ y = u \cdot v $, where $ u $ and $ v $ are functions of $ x $, then the derivative of $ y $ with respect to $ x $ is given by:

$$ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} $$

Here, $ u = x^2 $ and $ v = e^{-x} $. The derivatives of $ u $ and $ v $ with respect to $ x $ are:

$$ \frac{du}{dx} = 2x $$
$$ \frac{dv}{dx} = -e^{-x} $$

Now, applying the product rule:

$$ \frac{dy}{dx} = x^2 (-e^{-x}) + e^{-x} (2x) $$
$$ \frac{dy}{dx} = -x^2 e^{-x} + 2x e^{-x} $$
$$ \frac{dy}{dx} = e^{-x} (2x - x^2) $$

So, the derivative of $ y = x^2 e^{-x} $ with respect to $ x $ is $ e^{-x} (2x - x^2) $.