Anybody good with calculus product rule?
(x-2)e^(2-x)

- anonymous

Anybody good with calculus product rule?
(x-2)e^(2-x)

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- jamiebookeater

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- Mertsj

\[(x-2)(e ^{2-x})(-1)+e ^{2-x}(1)=(e ^{2-x})(2-x)\]

- Mertsj

Assuming that you want the first derivative.

- anonymous

Where does the (-1) come from??

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## More answers

- .Sam.

comes from chain rule.

- TuringTest

what is the derivative of\[2-x\]?

- anonymous

But I thought e differentiates to itself?

- TuringTest

but you need to use the chain rule as .Sam. said....

- TuringTest

\[{d\over dx}e^{2-x}=e^{2-x}\cdot{d\over dx}(2-x)=-e^{2-x}\]

- TuringTest

@shallster are you familiar with the chain rule?
or is the post above confusing you?

- anonymous

pretty confusing.

- TuringTest

ok, can you take the derivative of\[(2-x)^2\]?

- 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

yes, but you seem to have trouble differentiating\[e^{2-x}\]so I'm trying to help you do so

- anonymous

but i thought e differentiates to itself?

- 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

You obviously know what you're talking about, so I'm all ears. I NEED help.

- 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

4x(1-x^2) ???

- 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

\[\frac d{dx}e^{f(x)}=e^{f(x)}\cdot f'(x)\]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

do you agree with the above line?
let me know what is confusing you if not

- anonymous

so e^(2-x) = -1e^(2-x)

- TuringTest

yep :)
now does mertsj's answer make sense?

- 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

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

I also struggle with factoring. I'm such a douche.

- 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

Understood.

- 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

That I will.

- TuringTest

thanks y'all !
you guys make the site what it is though
agape love =)

- Mertsj

Right back to you, Turing.

- anonymous

Still not sure how the -2 became +2 when factoring?

- Mertsj

By distributing the -1

- 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

I understand the rest but how does (1-(x-2)) expand to (1-x+2)?

- TuringTest

distribute -1

- anonymous

I bet it's something really simple that my brain is bypassing!!

- TuringTest

like that^ ?
:)

- TuringTest

\[-(x-2)=-x+2\]no?

- anonymous

ah, i see. Thanks guys.

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