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

  • calculus -

    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
    y = -x

  • calculus -

    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...

  • calculus -

    sure it is normal. slope = -1/slope of function at origin

  • calculus -

    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
    du/dx = 1
    dv/dx = e^x
    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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