anonymous
  • anonymous
Anybody good with calculus product rule? (x-2)e^(2-x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Mertsj
  • Mertsj
\[(x-2)(e ^{2-x})(-1)+e ^{2-x}(1)=(e ^{2-x})(2-x)\]
Mertsj
  • Mertsj
Assuming that you want the first derivative.
anonymous
  • anonymous
Where does the (-1) come from??

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.Sam.
  • .Sam.
comes from chain rule.
TuringTest
  • TuringTest
what is the derivative of\[2-x\]?
anonymous
  • anonymous
But I thought e differentiates to itself?
TuringTest
  • TuringTest
but you need to use the chain rule as .Sam. said....
TuringTest
  • TuringTest
\[{d\over dx}e^{2-x}=e^{2-x}\cdot{d\over dx}(2-x)=-e^{2-x}\]
TuringTest
  • TuringTest
@shallster are you familiar with the chain rule? or is the post above confusing you?
anonymous
  • anonymous
pretty confusing.
TuringTest
  • TuringTest
ok, can you take the derivative of\[(2-x)^2\]?
anonymous
  • anonymous
it's not asking me to do that though?? The equation is (x-2)e^(2-x) I need the derivatives of x-2 & e^(2-x) don't I??
TuringTest
  • TuringTest
yes, but you seem to have trouble differentiating\[e^{2-x}\]so I'm trying to help you do so
anonymous
  • anonymous
but i thought e differentiates to itself?
TuringTest
  • TuringTest
again: you \(must\) use the chain rule here \[\frac d{dx}e^x=e^x\]yes, but this is not the case here, we have a compound function so we must use the chain rule
anonymous
  • anonymous
You obviously know what you're talking about, so I'm all ears. I NEED help.
TuringTest
  • TuringTest
so again I ask you to demonstrate your understanding of the chain rule what if I asked you to differentiate\[(1-x^2)^2\]by using the chain rule, could you do that?
anonymous
  • anonymous
4x(1-x^2) ???
TuringTest
  • TuringTest
exactly, you had to differentiate the inside function as well now that I know you understand this I can explain the other derivative...
TuringTest
  • TuringTest
\[\frac d{dx}e^{f(x)}=e^{f(x)}\cdot f'(x)\]see? you still leave the e^(whatever) part the same, but you still must differentiate the second function, as always in the chain rule
TuringTest
  • TuringTest
do you agree with the above line? let me know what is confusing you if not
anonymous
  • anonymous
so e^(2-x) = -1e^(2-x)
TuringTest
  • TuringTest
yep :) now does mertsj's answer make sense?
anonymous
  • anonymous
I've got it down to e^2-x) - (x-2)e^(2-x) and I'm not that sure how to simplify it.
TuringTest
  • TuringTest
dang lag... factor out \(e^{2-x}\) like so\[(x-2)e^{2-x}\]\[e^{2-x} -(x-2)e^{2-x}=e^{2-x}(1-x+2)=e^{2-x}(3-x)\]so I think mertsj was a little bit off
anonymous
  • anonymous
I also struggle with factoring. I'm such a douche.
TuringTest
  • TuringTest
no, this kind of factoring is a bit tricky I learned it through problems like these as you can see, mertsj made a small mistake in factoring as well, despite his illustrious abilities it happens to the best of us ;) PS: as a mod, I have to ask you not to say the "d" word above I'd like to keep you around after all!
anonymous
  • anonymous
Understood.
Mertsj
  • Mertsj
I can't believe I did that. But, shallster, as you will learn, Turing is the resident expert. So always pay attention to him.
anonymous
  • anonymous
That I will.
TuringTest
  • TuringTest
thanks y'all ! you guys make the site what it is though agape love =)
Mertsj
  • Mertsj
Right back to you, Turing.
anonymous
  • anonymous
Still not sure how the -2 became +2 when factoring?
Mertsj
  • Mertsj
By distributing the -1
Mertsj
  • Mertsj
\[e ^{2-x}-(x-2)e ^{2-x}=e ^{2-x}[1-(x-2)]=e ^{2-x}[1-x+2]=e ^{2-x}[3-x]\]
anonymous
  • anonymous
I understand the rest but how does (1-(x-2)) expand to (1-x+2)?
TuringTest
  • TuringTest
distribute -1
anonymous
  • anonymous
I bet it's something really simple that my brain is bypassing!!
TuringTest
  • TuringTest
like that^ ? :)
TuringTest
  • TuringTest
\[-(x-2)=-x+2\]no?
anonymous
  • anonymous
ah, i see. Thanks guys.

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