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Math Help - Derivatives and tangents

  1. #1
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    Derivatives and tangents

    Hi,

    Q1. find equation of the tangent line and the normal line to the curve
    y=(3x-2)e^-x at the point (0,-2).

    Q2. Find derivative of cos(x+y) + y = 0

    thanks,

    moon
    Last edited by Moon Hoplite; January 6th 2009 at 07:17 AM. Reason: wrong title
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  2. #2
    Super Member craig's Avatar
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    What experience of differentiation have you got?

    You need some experience with this to solve these types of questions.
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  3. #3
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    Quote Originally Posted by Moon Hoplite View Post
    Hi,

    Q1. find equation of the tangent line and the normal line to the curve
    y=(3x-2)e^-x at the point (0,-2).

    Q2. Find derivative of cos(x+y) + y = 0

    thanks,

    moon
    to Q1:

    1, Use product and chain rule to calculate the first derivative ( y'=(5-3x)e^{-x})

    2. Plug in x = 0 to get the slope of the tangent (m = 5)

    3. Use point-slope-formula of the straight line to get the equation of the tangent (y+2=5(x-0) )

    4. Calculate the negative reciprocal of m to get the slope of the normal to the curve in (0, -2): ( m_\perp = -\dfrac15 )

    5. Use point-slope-formula to get the equation of the normal: ( y+2 = -\dfrac15 (x-0) )

    6. Now it's your turn!
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  4. #4
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    hi thanks, i know how to find derivatives, but i dont know how to expand
    "cos(x+y) + y = 0" in terms of y, could anyone help on that question?
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  5. #5
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    Quote Originally Posted by Moon Hoplite View Post
    Hi,

    Q1. find equation of the tangent line and the normal line to the curve
    y=(3x-2)e^-x at the point (0,-2).

    Q2. Find derivative of cos(x+y) + y = 0

    thanks,

    moon
    to Q2:

    This is an implicite function:

    f(x,y) = \cos(x+y)+y=0

    You are asked to calculate the drivative:

    \dfrac{dy}{dx}=y'= - \dfrac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}

    \dfrac{\partial f}{\partial x} = -\sin(x+y)

    \dfrac{\partial f}{\partial y} = 1-\sin(x+y)

    Therefore:

    \dfrac{dy}{dx}=y'=  \dfrac{\sin(x+y)}{1-\sin(x+y)}
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