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

- anonymous

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

- jamiebookeater

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- anonymous

@Callisto

- anonymous

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

- anonymous

yeah it is

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## More answers

- anonymous

\[g(x)=e^{\frac{1}{x}}\]
\[g'(x)=-\frac{1}{x^2}e^{\frac{1}{x}}\] by the chain rule

- anonymous

so the second part of your product rule is the problem. first part is good

- anonymous

but how is it (-1/x^2) e^(1/x)?

- anonymous

shouldn't the derivative e^(1/x)

- Callisto

\[y=e^{\frac{1}{x}}\]
Let u = 1/x
\[\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx} = \frac{d}{du}e^u \times \frac{d}{dx}(\frac{1}{x})= ...?\]

- anonymous

1?

- anonymous

or 0

- anonymous

by quotient rule i did (1)'(x)-(1)(x)'/(x)^@

- Callisto

You can use power for d/dx (1/x)
\[\frac{d}{dx} \frac{1}{x} = \frac{d}{dx} (x^{-1}) = (-1)x^{-1-1} =...?\]

- anonymous

-x^(-2)?

- Callisto

Yes.

- anonymous

ok so it will be e^(-x^(-2))?

- Callisto

The power of e would not change when you differentiate e^(something)!

- Callisto

\[\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx} = \frac{d}{du}e^u \times \frac{d}{dx}(\frac{1}{x})= ...?\]
u = 1/x and you found d/dx(1/x). So......

- anonymous

i'm lost with this one

- Callisto

Where?

- anonymous

are you asking for the derivative of (1/x) above? i dont understand what exactly you are asking

- Callisto

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

Is that clear? Do you understand what we are working on?

- anonymous

but isn't derivative of (1/x) = -x^(-2) ? i thought we got that right

- anonymous

wait. give me second

- Callisto

Yes. d/dx (1/x) = -x^(-2) = -1/x^2

- anonymous

so you are asking derivative of e^(1/x) * -1/x^2?

- Callisto

No. Instead, derivative of e^(1/x), which is equal to e^(1/x) * -1/x^2. Do you understand how to get it?

- anonymous

so looking at the original question question my answer should be (2x)(e^1/x)-(x^2)(e^(1/x))?

- anonymous

derivative of (x^(2))(e^(1/x)) is the original question

- Callisto

No... The second part is not correct.
For the second part, you need to find the derivative of e^(1/x). What is it?

- anonymous

isn't it e^(1/x)

- anonymous

e^(1/x) * -1/x^2?

- Callisto

It is *NOT* e^(1/x)
e^(1/x) * -1/x^2
^How do you get it?

- anonymous

because derivative of 1/x = -1/x^2

- Callisto

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

^That is your derivative btw.

- anonymous

yes i had already gotten the first part. just the second was the problem

- Callisto

Just show us what you've got.

- anonymous

final answer for the original question is (2x)(e^(1/x))-(x^2)(1/x^2)(e^(1/x))

- anonymous

this is what i got

- Callisto

That looks pretty cool~ But you can simplify the last term.

- anonymous

i don't think we have to. but you can show me if you want

- Callisto

\[(2x)(e^{\frac{1}{x}})-(x^2)\frac{1}{x^2}(e^{\frac{1}{x}}) = (2x)(e^{\frac{1}{x}})-\frac{x^2}{x^2}(e^{\frac{1}{x}}) =...?\]

- anonymous

x^2 / x^2 gets cancelled out = 1

- Callisto

Yes.

- anonymous

perfect.

- anonymous

Thank you very much for your help

- anonymous

you explain concepts very well

- Callisto

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 :)

- anonymous

No, not confusing at all

- anonymous

yes that looks better

- Callisto

:)

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