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Math Help - Implicit differentiation and tangent lines

  1. #1
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    Implicit differentiation and tangent lines

    One of these I know is wrong ,one of these I don't understand how to do.

    1) Use logarithmic differentiation to find the derivative of the function.
    ln(y) = ln(x)ln(sin(x))
    F(x) = ln(x)ln(sin(x))
    f(x) = ln(x); f'(x) = 1/x
    g(x) = ln(sin(x))
    g'(x) = cosx/sinx

    y'/y = [ln(sin(x))/x] + [ln(x)cos(x)/sin(x)]
    y' = y{[ln(sin(x))/x] + [ln(x)cos(x)/sin(x)]}

    y' = [sin(x)^ln(x)ln(sin(x))/x] + (ln(x)^2) * cos(x)

    Not sure where i went wrong on that one.

    2) The curve with equation y2 = x3 + 3x2 is called the Tschirnhausen cubic. -Find an equation of the tangent line to this curve at the point (1, -2).
    -Find the points on the curve where the tangent line has a vertical asymptote

    I was under the impression I had to derive the function, and then find points where it is undefined, but the question is asking for y, not y'.
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  2. #2
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    Quote Originally Posted by CFem View Post
    One of these I know is wrong ,one of these I don't understand how to do.

    1) Use logarithmic differentiation to find the derivative of the function.
    ln(y) = ln(x)ln(sin(x))
    F(x) = ln(x)ln(sin(x))
    f(x) = ln(x); f'(x) = 1/x
    g(x) = ln(sin(x))
    g'(x) = cosx/sinx

    y'/y = [ln(sin(x))/x] + [ln(x)cos(x)/sin(x)]
    y' = y{[ln(sin(x))/x] + [ln(x)cos(x)/sin(x)]}

    y' = [sin(x)^ln(x)ln(sin(x))/x] + (ln(x)^2) * cos(x)

    Not sure where i went wrong on that one.

    2) The curve with equation y2 = x3 + 3x2 is called the Tschirnhausen cubic. -Find an equation of the tangent line to this curve at the point (1, -2).
    -Find the points on the curve where the tangent line has a vertical asymptote ... recheck this statment

    I was under the impression I had to derive the function, and then find points where it is undefined, but the question is asking for y, not y'.
     <br />
y = (\sin{x})^{\ln{x}}<br />

    \ln{y} = \ln\left[(\sin{x})^{\ln{x}}\right]

     <br />
\ln{y} = \ln{x} \cdot \ln(\sin{x})<br />

     <br />
\frac{y'}{y} = \ln{x} \cdot \cot{x} + \ln(\sin{x}) \cdot \frac{1}{x}<br />

     <br />
y' = (\sin{x})^{\ln{x}}\left[\ln{x} \cdot \cot{x} + \ln(\sin{x}) \cdot \frac{1}{x}\right]<br />


     <br />
y^2 = x^3 + 3x^2<br />

     <br />
2y \cdot y' = 3x^2 + 6x<br />

     <br />
y' = \frac{3x(x+2)}{2y}<br />

    at (1,-2) , m = -\frac{9}{4}

     <br />
y + 2 = -\frac{9}{4}(x - 1)<br />

    tangent lines are vertical when y' is undefined ...

    y' = \frac{3x(x+2)}{2y} is undefined when y = 0

    points on the curve where y = 0 ...

     <br />
0 = x^3 + 3x^2<br />

     <br />
0 = x^2(x + 3)<br />

    x = 0 , x = -3

    points where the tangent line is vertical are (0,0) and (-3,0)
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  3. #3
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    #2 is wrong.

    Thanks for the help so far, though.

    Edit: And I was wrong, the question should read "The curve with equation y2 = x3 + 3x2 is called the Tschirnhausen cubic. Find an equation of the tangent line to this curve at the point (1, -2)

    At what points does this curve have a horizontal tangent?"

    However both parts were still maked incorrect, and the points are supposed to have different y values.
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  4. #4
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    Quote Originally Posted by CFem View Post

    Edit: And I was wrong, the question should read "The curve with equation y^2 = x^3 + 3x^2 is called the Tschirnhausen cubic. Find an equation of the tangent line to this curve at the point (1, -2)

    At what points does this curve have a horizontal tangent?"
    first, note how to denote exponents in your posts by looking at the correction above.



    the derivative is the same as completed above.

    substitute the given coordinates (1,-2) into the derivative to determine the slope of the tangent line ... then use the point-slope form of a linear equation to find the tangent line equation.


    a horizontal tangent line occurs where \frac{dy}{dx} = 0

    set the derivative equal to 0, solve for x in terms of y or y in terms of x (whichever is easier), and substitute the result into the original equation for the curve ... solve.
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  5. #5
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    Sorry about the notation. Was copy and pasting.

    y = (-9/4x) -1/4

    Solving dy/dx for x gave me -2 (marked correct) and 0 instead of -3 and 0.

    Which gave me the points:

    (0,-1/4)
    (-2,17/4)

    Am I mistaken? Still is getting refused (except for the -2)

    Edit: Nevermind, I lost a sign. I got (-9/4)x + (1/4)

    But the only coordinate that isn't being marked off is -2. -2 in that equation gives 19/4, correct? It's also still refusing to accept 0 as a valid point.

    Edi2: I don't think I'm using the right equation. It doesn't look like subbing the x values into y' gives the proper number. Should the number be substituted into the original equation?
    Last edited by CFem; October 8th 2009 at 04:26 PM.
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