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Math Help - tangent line

  1. #1
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    tangent line

    find the equation of the tangent line to y=(sin x)^4x at the point (pi/2, 1)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Bust2000 View Post
    find the equation of the tangent line to y=(sin x)^4x at the point (pi/2, 1)
    a line is of the form y = mx + b

    we will find the line by the point-slope form. that is, y - y_1 = m(x - x_1), where m is the slope (given by the derivative evaluated at (x_1,y_1)) and (x_1,y_1) is a point the slope passes through (here it is (pi/2, 1)).

    can you continue?
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  3. #3
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    Quote Originally Posted by Bust2000 View Post
    find the equation of the tangent line to y=(sin x)^4x at the point (pi/2, 1)
    Do you really mean:

    f(x) = \left(\sin(x)\right)^{4x} ...... ??

    If so re-write f to:

    f(x) = e^{4x \cdot \ln(\sin(x))} ...... and use the chain rule to calculate the derivative:

    f'(x) = \frac{e^{4x \cdot \ln(\sin(x))}}{\sin(x)} \cdot \left(4 \sin(x) \cdot \ln(\sin(x)) + 4x \cdot \cos(x)  \right)

    Now plug in x = \frac{\pi}{2} to calculate the slope of the tangent.

    Since \ln\left(\sin\left(\frac{\pi}2 \right) \right) = 0 and \cos \left(\frac{\pi}2 \right) = 0 too you'll get:

    f'\left(\frac{\pi}2 \right) = 0 and therefore the equation of the tangent is

    y = 0

    EDIT: According to Jhevons reply the equation of the tangent is y = 1
    Last edited by earboth; February 11th 2008 at 12:17 AM.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by earboth View Post
    Do you really mean:

    f(x) = \left(\sin(x)\right)^{4x} ...... ??

    If so re-write f to:

    f(x) = e^{4x \cdot \ln(\sin(x))} ...... and use the chain rule to calculate the derivative:

    f'(x) = \frac{e^{4x \cdot \ln(\sin(x))}}{\sin(x)} \cdot \left(4 \sin(x) \cdot \ln(\sin(x)) + 4x \cdot \cos(x)  \right)

    Now plug in x = \frac{\pi}{2} to calculate the slope of the tangent.

    Since \ln\left(\sin\left(\frac{\pi}2 \right) \right) = 0 and \cos \left(\frac{\pi}2 \right) = 0 too you'll get:

    f'\left(\frac{\pi}2 \right) = 0 and therefore the equation of the tangent is

    y = 0
    i believe it would be y = 1 if the slope is zero
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