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derivative of (x^(2))(e^(1/x))

Mathematics
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i got (2x)(e^(1/x))+x^(2)(e^(1/x) which is wrong
yeah it is

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Other answers:

\[g(x)=e^{\frac{1}{x}}\] \[g'(x)=-\frac{1}{x^2}e^{\frac{1}{x}}\] by the chain rule
so the second part of your product rule is the problem. first part is good
but how is it (-1/x^2) e^(1/x)?
shouldn't the derivative e^(1/x)
\[y=e^{\frac{1}{x}}\] Let u = 1/x \[\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx} = \frac{d}{du}e^u \times \frac{d}{dx}(\frac{1}{x})= ...?\]
1?
or 0
by quotient rule i did (1)'(x)-(1)(x)'/(x)^@
You can use power for d/dx (1/x) \[\frac{d}{dx} \frac{1}{x} = \frac{d}{dx} (x^{-1}) = (-1)x^{-1-1} =...?\]
-x^(-2)?
Yes.
ok so it will be e^(-x^(-2))?
The power of e would not change when you differentiate e^(something)!
\[\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx} = \frac{d}{du}e^u \times \frac{d}{dx}(\frac{1}{x})= ...?\] u = 1/x and you found d/dx(1/x). So......
i'm lost with this one
Where?
are you asking for the derivative of (1/x) above? i dont understand what exactly you are asking
I asked derivative of 1/x because you didn't get that right. To get the derivative of derivative of (x^(2))(e^(1/x)), it should be like this: \[y = x^2e^{\frac{1}{x}}\]\[y' = x^2\frac{d}{dx}e^{\frac{1}{x}} + e^{\frac{1}{x}}\frac{d}{dx}(x^2)\] Then, then you need to work out what d/dx (e^(1/x)) is and d/dx (x^2) are. The later one is easy, and you got that right. The problem is to find d/dx (e^(1/x)) So, let u = 1/x \[\frac{d}{dx}e^{\frac{1}{x}} = \frac{d}{du}(e^u) \times \frac{d}{dx} (\frac{1}{x}) = ...?\]
Is that clear? Do you understand what we are working on?
but isn't derivative of (1/x) = -x^(-2) ? i thought we got that right
wait. give me second
Yes. d/dx (1/x) = -x^(-2) = -1/x^2
so you are asking derivative of e^(1/x) * -1/x^2?
No. Instead, derivative of e^(1/x), which is equal to e^(1/x) * -1/x^2. Do you understand how to get it?
so looking at the original question question my answer should be (2x)(e^1/x)-(x^2)(e^(1/x))?
derivative of (x^(2))(e^(1/x)) is the original question
No... The second part is not correct. For the second part, you need to find the derivative of e^(1/x). What is it?
isn't it e^(1/x)
e^(1/x) * -1/x^2?
It is *NOT* e^(1/x) e^(1/x) * -1/x^2 ^How do you get it?
because derivative of 1/x = -1/x^2
Yup.. So, can you solve \(y' = x^2\frac{d}{dx}e^{\frac{1}{x}} + e^{\frac{1}{x}}\frac{d}{dx}(x^2)\) now?
^That is your derivative btw.
yes i had already gotten the first part. just the second was the problem
Just show us what you've got.
final answer for the original question is (2x)(e^(1/x))-(x^2)(1/x^2)(e^(1/x))
this is what i got
That looks pretty cool~ But you can simplify the last term.
i don't think we have to. but you can show me if you want
\[(2x)(e^{\frac{1}{x}})-(x^2)\frac{1}{x^2}(e^{\frac{1}{x}}) = (2x)(e^{\frac{1}{x}})-\frac{x^2}{x^2}(e^{\frac{1}{x}}) =...?\]
x^2 / x^2 gets cancelled out = 1
Yes.
perfect.
Thank you very much for your help
you explain concepts very well
You're welcome. I hope I didn't confuse you :S \[(2x)(e^{\frac{1}{x}})-\frac{x^2}{x^2}(e^{\frac{1}{x}}) =(2x)(e^{\frac{1}{x}})-e^{\frac{1}{x}}=e^{\frac{1}{x}}(2x-1)\]That looks nice :)
No, not confusing at all
yes that looks better
:)

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