Question

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

Answer 1

`dy/dx=xe^-x(2-x).`

Explanation:

`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).`

Answer 2

`xe^(-x)(2-x)`

Explanation:

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)`