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Math Help - tricky derivatives

  1. #1
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    tricky derivatives

    1. Let f(x)= -x^3 - 2x^2 + x + 1 and g(x)= ln(x+1)+1
    a. Find the equation of the line tangent to f at x=0
    b. Show that g has the same tangent line as f at x=0


    2. Given x=t^4 - t^2 + t and y=2t^3 - t, find dy/dx

    3. Given x^2 + 3y^2 = 1
    a. Find y'
    b. Find the slope of the curve at the point (.5,.5)
    c. Find the equation of the line tangent to the ellipse at (.5,.5)
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  2. #2
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    Quote Originally Posted by Dr. Noobles View Post
    2. Given x=t^4 - t^2 + t and y=2t^3 - t, find dy/dx
    You find the parameter derivative,
    \frac{dy}{dt}=6t^2-1

    And you find the parameter derivative,
    \frac{dx}{dt}=4t^3-2t+1

    Thus,
    \frac{dy}{dx}=\frac{6t^2-1}{4t^3-2t+1}
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  3. #3
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    Quote Originally Posted by Dr. Noobles View Post
    1. Let f(x)= -x^3 - 2x^2 + x + 1 and g(x)= ln(x+1)+1
    a. Find the equation of the line tangent to f at x=0
    b. Show that g has the same tangent line as f at x=0
    a) f'(x) = -3x^2 - 4x + 1

    So the slope of the tangent line at x = 0 is:
    m = f'(0) = 1

    The tangent line is thus: y = x + b where b is the intercept. This line touches the function y = f(x), and we know that y = f(0) = 1. So the line y = x + b passes through the point (0, 1). Thus
    1 = 0 + b gives b = 1.

    Thus your tangent line is y = x + 1. (See graph below.)

    b) To show that g(x) has the same tangent line at x = 0, all we need to do is verify that g(0) = 1 and that g'(0) = 1.

    So:
    g(0) = ln(x+1) + 1 = ln(0 + 1) + 1 = ln(1) + 1 = 0 + 1 = 1 (Check!)
    g'(x) = \frac{1}{x+1}
    so
    g'(x) = \frac{1}{0 + 1} = 1 (Check!)

    -Dan
    Attached Thumbnails Attached Thumbnails tricky derivatives-f-x-.jpg   tricky derivatives-g-x-.jpg  
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Dr. Noobles View Post
    3. Given x^2 + 3y^2 = 1
    a. Find y'
    b. Find the slope of the curve at the point (.5,.5)
    c. Find the equation of the line tangent to the ellipse at (.5,.5)
    a) Use implicit differentiation:
    x^2 + 3y^2 = 1

    2x + 6yy' = 0

    6yy' = -2x

    y' = -\frac{x}{3y}

    b) At \left ( \frac{1}{2}, \frac{1}{2} \right ):

    y' = -\frac{\frac{1}{2}}{3\frac{1}{2}} = -\frac{1}{3}

    c) We know the slope is m = -\frac{1}{3} and the tangent line passes through the point \left ( \frac{1}{2}, \frac{1}{2} \right ) so:
    y = mx + b

    \frac{1}{2} = -\frac{1}{3} \cdot \frac{1}{2} + b

    b = \frac{1}{2} + \frac{1}{6} = \frac{2}{3}

    So the tangent line is
    y = -\frac{1}{3}x + \frac{2}{3}

    -Dan
    Attached Thumbnails Attached Thumbnails tricky derivatives-ellipse.jpg  
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