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Thread: equztion of tangent

  1. #1
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    equztion of tangent

    (a) Find equation of the tangent line and the normal line to the curve
    y
    = (3x -2)e^-xat the point (0;-2);
    (b) Find
    dy/dx given that cos(x + y) + y = 0.


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  2. #2
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    Quote Originally Posted by trythis View Post
    (a) Find equation of the tangent line and the normal line to the curve
    y
    = (3x -2)e^-xat the point (0;-2);
    (b) Find
    dy/dx given that cos(x + y) + y = 0.


    Tangents and normals
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  3. #3
    Member Greengoblin's Avatar
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    1.)

    $\displaystyle y=(3x-2)e^{-x}$

    $\displaystyle \frac{dy}{dx}=3e^{-x}-(3x-2)e^{-x}=(5-3x)e^{-x}$

    $\displaystyle y'(0)=5$

    $\displaystyle y-y_1=m(x-x_1)$

    $\displaystyle y+2=5(x-0)$

    $\displaystyle y=5x-2$

    $\displaystyle m_1=-\frac{1}{m}=-\frac{1}{5}$

    $\displaystyle y-y_1=m_1(x-x_1)$

    $\displaystyle y+2=-\frac{1}{5}(x-0)$

    $\displaystyle y=-\frac{x}{5}-2$

    2.)

    $\displaystyle \cos(x+y)+y=0$

    $\displaystyle y=-\cos(x+y)$

    $\displaystyle y'=\frac{d}{dx}(-\cos(x+y))$

    $\displaystyle y'=(1+y')\sin(x+y)=\sin(x+y)+y'\sin(x+y)$

    $\displaystyle y'-y'\sin(x+y)=\sin(x+y)$

    $\displaystyle y'[1-\sin(x+y)]=\sin(x+y)$

    $\displaystyle y'=\frac{\sin(x+y)}{1-\sin(x+y)}$
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