Working for product rule and chain rule questions.

[Whatlol]

let y= (x^2 - 2x)e^x

find dy/dx

dy/dx = 2xe^(x) + x^2e^x - (2e^x +2xe^(x))
        = e^x(x^2 -2)

For two marks is that sufficient working, or do i need to state im using product rule and actually show the application of the rule.

Thanks for the help.

[kenhung123]

define u=(x^2-2x) and v=e^2x

dy/dx=vdu/dx+udv/dx
       =vu'+uv'
       =e^2x(2x-2)+2e^2x(x^2-2x)

Perhaps state "Using chain rule": working, QED

[costa94]

Might want to get your differentiation right lol.
It should be dy/dx = (x^2 - x - 1)2e^2x, if I read your y right as y= (x^2 - 2x)e^2

[Whatlol]

Might want to get your differentiation right lol.
It should be dy/dx = (x^2 - x - 1)2e^2x, if I read your y right as y= (x^2 - 2x)e^2

sorry... it should be (x^2 - 2x)e^x

[costa94]

Ok in that case it's e^x(x^2 - 2) as you have
Basically for two marks you need a line of working which clearly shows product rule use, thn the answer.

[TrueTears]

let y= (x^2 - 2x)e^x

find dy/dx

dy/dx = 2xe^(x) + x^2e^x - (2e^x +2xe^(x))
        = e^x(x^2 -2)

For two marks is that sufficient working, or do i need to state im using product rule and actually show the application of the rule.

Thanks for the help.

perfectly sufficient, your first line shows the application, 2nd line is answer mark. done!

[Whatlol]

let y= (x^2 - 2x)e^x

find dy/dx

dy/dx = 2xe^(x) + x^2e^x - (2e^x +2xe^(x))
        = e^x(x^2 -2)

For two marks is that sufficient working, or do i need to state im using product rule and actually show the application of the rule.

Thanks for the help.

perfectly sufficient, your first line shows the application, 2nd line is answer mark. done!

oh really! awesome i was thinking i had to write out the rule and everything.

Thanks for that TT +karma