Another calculating y'

• Jan 3rd 2010, 01:45 PM
spazzyskylar
Another calculating y'
If y=xe^(-1/x)

is y'= xe^(-1/x)*(-(x-1)/x^2)+e^(-1/x)
• Jan 3rd 2010, 01:51 PM
TWiX
Quote:

Originally Posted by spazzyskylar
If y=xe^(-1/x)

is y'= xe^(-1/x)*(-(x-1)/x^2)+e^(-1/x)

You said
d/dx(-1/x) = -(x-1) / x^2

• Jan 3rd 2010, 01:53 PM
skeeter
Quote:

Originally Posted by spazzyskylar
If y=xe^(-1/x)

is y'= xe^(-1/x)*(-(x-1)/x^2)+e^(-1/x)

?

note that the derivative of $\displaystyle -\frac{1}{x}$ is $\displaystyle \frac{1}{x^2}$
• Jan 3rd 2010, 01:55 PM
spazzyskylar
Not really, should it be -(x+1)/x^2
• Jan 3rd 2010, 01:56 PM
Not quite! if $\displaystyle u=-\frac{1}{x}$

$\displaystyle \frac{d}{dx}(xe^\frac{-1}{x})=\frac{d}{dx}xe^u=x\frac{d}{dx}e^u+e^u(1)$

$\displaystyle =x\frac{du}{du}\frac{d}{dx}e^u+e^u=x\frac{du}{dx}\ frac{d}{du}e^u+e^u$

$\displaystyle =(x)\frac{1}{x^2}e^u+e^u=\frac{e^\frac{-1}{x}}{x}+e^\frac{-1}{x}$
• Jan 3rd 2010, 01:59 PM
e^(i*pi)
Quote:

Originally Posted by spazzyskylar
If y=xe^(-1/x)

is y'= xe^(-1/x)*(-(x-1)/x^2)+e^(-1/x)

$\displaystyle u = x \: \rightarrow \: u' = 1$

$\displaystyle v = e^{-\frac{1}{x}} \: \: \rightarrow \: v' = \frac{1}{x^2}e^{-\frac{1}{x}}$

skeeter gave the derivative of the exponent so multiply due to the chain rule

Now use the product rule

y' = u'v + uv'

Spoiler:
$\displaystyle e^{-\frac{1}{x}} + \frac{1}{x}e^{-\frac{1}{x}} = e^{-\frac{1}{x}}\left(1+\frac{1}{x}\right)$
• Jan 3rd 2010, 02:00 PM
spazzyskylar
I got it now :)
• Jan 3rd 2010, 02:05 PM
pickslides
$\displaystyle y'=x'e^{-1\over x} +x(e^{-1\over x})'$