Another calculating y'

**spazzyskylar** · January 3rd 2010, 01:45 PM

If y=xe^(-1/x)

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

**TWiX** · January 3rd 2010, 01:51 PM

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

Are you sure about this?

**skeeter** · January 3rd 2010, 01:53 PM

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 $-\frac{1}{x}$ is $\frac{1}{x^2}$

**spazzyskylar** · January 3rd 2010, 01:55 PM

Not really, should it be -(x+1)/x^2

**Archie Meade** · January 3rd 2010, 01:56 PM

Not quite! if $u=-\frac{1}{x}$

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

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

$=(x)\frac{1}{x^2}e^u+e^u=\frac{e^\frac{-1}{x}}{x}+e^\frac{-1}{x}$

**e^(i*pi)** · January 3rd 2010, 01:59 PM

Originally Posted by spazzyskylar

If y=xe^(-1/x)

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

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

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

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

Now use the product rule

y' = u'v + uv'

Spoiler:

**spazzyskylar** · January 3rd 2010, 02:00 PM

I got it now

**pickslides** · January 3rd 2010, 02:05 PM

$y'=x'e^{-1\over x} +x(e^{-1\over x})'$

Thread: Another calculating y'

LinkBack

Thread Tools

Display

Another calculating y'

Similar Math Help Forum Discussions

Calculating mean

Calculating APR

Calculating y'

calculating P(X>=Y)

calculating p-value

Search Tags