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You need to use the quotient rule which says if y=u/v then y'=(u'v-uv')/v^2
In this case u=x^2 so u'=2x and v=e^x so v'=e^xSo y'=(2x.e^x-x^2.e^x)/(e^x)^2
After factorising for e^x at the denominator and numerator and eliminating it you get:
y'=(2x-x^2)/e^x. We need y' when x=0 so sub 0 for x to get y'=0.
Hope it is useful.
Reza Bokat,
Cambridge Coaching
www.cambridgecoaching.com.au
I don't think this was necessary at all.I think you need to revise your maths. did you use the quotient rule?
Whoever taught you that is working for the DEVIL! That rule only works when the power is a constant (in this case, 'x' is a variable). When u differentiate ex, the result is just ex (this is the easiest thing u need to differentiate, its just itself!!!!)Thanks heaps dude. But why didn't you do this, when you derive the v=e^x, why didn't you do v'=xe^x-1 ?
I was taught to do it that way --- helpppppp I have an exam tomorrow
Thanks in advanceee
e^x is the derivative of itself. In other words when you differentiate e^x, it does not change.Thanks heaps dude. But why didn't you do this, when you derive the v=e^x, why didn't you do v'=xe^x-1 ?
I was taught to do it that way --- helpppppp I have an exam tomorrow
Thanks in advanceee
It is a part of calculus, it's just not taught straight away during the differentiation chapter.Ohhhhh I think you learn this is integration or something? Because I haven't done that chapter yet. Sorry about that. I just saw it in past papers and thought it was part of calculus or whatever. Thanks everyone, really appreciate it![]()