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