Asked by sylvia
find an equation for a line that is normal to the graph y=e^x*x and goes through the origin.
Answers
Answered by
Damon
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
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
Answered by
sylvia
yeah that's 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...
Answered by
Damon
sure it is normal. slope = -1/slope of function at origin
Answered by
Damon
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
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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