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

find an equation for a line that is normal to the graph y=e^x*x and goes through the origin.

- calculus -
**Damon**, Monday, December 1, 2008 at 4:54pm
do 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 -
**Damon**, Monday, December 1, 2008 at 5:08pm
sure it is normal. slope = -1/slope of function at origin

- calculus -
**Damon**, Monday, December 1, 2008 at 5:13pm
Maybe 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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