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December 20, 2014

December 20, 2014

Posted by **sylvia** on Monday, December 1, 2008 at 3:34pm.

- calculus -
**Damon**, Monday, December 1, 2008 at 4:54pmdo we mean y = x e^x ???

at x = 0, y = 0 so it goes through the origin

slope = dy/dx = x e^x - e^x

at x = 0 that is e^0 = 1

so our slope = -1/1 = -1

so our line is

y = -1 x + 0

or

y = -x

- calculus -
**sylvia**, Monday, December 1, 2008 at 5:06pmyeah thats what i mean...but u cant plug in zero because it is not normal at the origin only goes through it..n the derivative is addition...

- calculus -
- calculus -
**Damon**, Monday, December 1, 2008 at 5:08pmsure it is normal. slope = -1/slope of function at origin

- calculus -
**Damon**, Monday, December 1, 2008 at 5:13pmMaybe you did not follow the derivative? I used the product rule

d/dx(uv) = u dv/dx + v du/dx

here u = x

and v = e^x

so

du/dx = 1

dv/dx = e^x

so

df/dx = x e^x + 1 e^x

at zero that is one because e^0 is one

so our function slope at the origin is m = 1

normal slope = -1/m = -1/1 = -1

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