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Math Help - derivatives

  1. #1
    Member >_<SHY_GUY>_<'s Avatar
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    Unhappy derivatives

    how do you get the derivative of:
    (\frac {1-\sin}{1+\cos})^2
    *hopefully i used the LaTex Does Correctly...

    i tried following my teacher in this problem but easily got lost...
    y = 2\sqrt{x-1} \csc^- \sqrt{x}

    and how would i find the normal and tangent line of:
    2x^2y + 2y^2=x^3 + 1 at (-1,1)?
    i know i have to take the derivative of the equation, but then what?
    Last edited by >_<SHY_GUY>_<; April 26th 2009 at 12:50 PM. Reason: i saw that my teacher corrected the problem
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  2. #2
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    how do you get the derivative of:
    (\frac {1-\sin}{1+\cos})^2
    *hopefully i used the LaTex Does Correctly...
    Hi

    Let u(x) = \frac {1-\sin x}{1+\cos x}

    The derivative of u^2 is 2\:u(x)\:u'(x)

    The quotient rule is used to calculate u'(x)

    1+\cos x) - (- \sin x)\1-\sin x)}{(1+\cos x)^2} = \frac{-1 -\cos x + \sin x}{(1+\cos x)^2}" alt="u'(x) = \frac{-\cos x\1+\cos x) - (- \sin x)\1-\sin x)}{(1+\cos x)^2} = \frac{-1 -\cos x + \sin x}{(1+\cos x)^2}" />

    The derivative of \left(\frac {1-\sin x}{1+\cos x}\right)^2 is therefore

    2\:\frac {1-\sin x}{1+\cos x}\:\:\frac{-1 -\cos x + \sin x}{(1+\cos x)^2}
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  3. #3
    Member >_<SHY_GUY>_<'s Avatar
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    Great Work...i messed up in accidentally ignoring the negative sign when taking the derivative of -\sin x
    thanks!

    Quote Originally Posted by running-gag View Post
    Hi

    Let u(x) = \frac {1-\sin x}{1+\cos x}

    The derivative of u^2 is 2\:u(x)\:u'(x)

    The quotient rule is used to calculate u'(x)

    1+\cos x) - (- \sin x)\1-\sin x)}{(1+\cos x)^2} = \frac{-1 -\cos x + \sin x}{(1+\cos x)^2}" alt="u'(x) = \frac{-\cos x\1+\cos x) - (- \sin x)\1-\sin x)}{(1+\cos x)^2} = \frac{-1 -\cos x + \sin x}{(1+\cos x)^2}" />

    The derivative of \left(\frac {1-\sin x}{1+\cos x}\right)^2 is therefore

    2\:\frac {1-\sin x}{1+\cos x}\:\:\frac{-1 -\cos x + \sin x}{(1+\cos x)^2}
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  4. #4
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    To your second question, the derivative of your original function gives you the slope of the line tangent to that point.

    Your tangent line will be in the form y = ax + b. Where a is the slope you got from your derivative at (-1, 1)

    Say that slope was 2. Now, to calculate b, use the point you were given.

    1 = 2(-1) + b

    b = 3

    So you're tangent line would be y = 2x + 3.

    And the normal line is just the inverse of the tangent.
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  5. #5
    Member >_<SHY_GUY>_<'s Avatar
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    then i wouldnt have to worry about 2y^2 being 4yy prime?

    Quote Originally Posted by derfleurer View Post
    To your second question, the derivative of your original function gives you the slope of the line tangent to that point.

    Your tangent line will be in the form y = ax + b. Where a is the slope you got from your derivative at (-1, 1)

    Say that slope was 2. Now, to calculate b, use the point you were given.

    1 = 2(-1) + b

    b = 3

    So you're tangent line would be y = 2x + 3.

    And the normal line is just the inverse of the tangent.
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  6. #6
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    Nope. You only use the derivative to find the slope at that point. After that, you're only interested in the original function.

    2x^2y + 2y^2=x^3

    For the slope, derive implicitly

    4xy + 2x^2y' + 4yy' = 3x^2

    y'(2x^2 + 4y) = 3x^2 - 4xy

    y' = \frac{3x^2 - 4xy}{2x^2 + 4y}


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  7. #7
    Member >_<SHY_GUY>_<'s Avatar
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    ok, now i see my mistake...i didnt know that the product rule applied to 2x^2yi see why i get some problems wrong...thank you!

    Quote Originally Posted by derfleurer View Post
    Nope. You only use the derivative to find the slope at that point. After that, you're only interested in the original function.

    2x^2y + 2y^2=x^3

    For the slope, derive implicitly

    4xy + 2x^2y' + 4yy' = 3x^2

    y'(2x^2 + 4y) = 3x^2 - 4xy

    y' = \frac{3x^2 - 4xy}{2x^2 + 4y}

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  8. #8
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    A lot of people slip up on implicit differentiation. If you want a tip

    Rewrite the question so that everything is on one side of the equals sign.

    2x^2y + 2y^2=x^3

    becomes

    2x^2y + 2y^2 - x^3=0

    Now consider this its own function.

    z = 2x^2y + 2y^2 - x^3

    Take the derivative of z with respect to x. This means considering all the y as just constants. So just like the derivative of 4x^2 yeilds 8x, the derivative of 4x^2y yields 8xy:

    \frac{dz}{dx} = 4xy - 3x^2

    Now do the same for y

    \frac{dz}{dy} = 2x^2 + 4y

    Now your derivative will be in the form \frac{-(dz/dx)}{dz/dy}

    \frac{-(4xy - 3x^2)}{2x^2 + 4y}
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  9. #9
    Member >_<SHY_GUY>_<'s Avatar
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    id say thats a really good trick there...thank you for that...i really appreciate it

    ok, so i substitute (-1,-1) in the derivative of the equation to get the slope, then in the original equation for the b-intercept or how does that work? i get confused in finding the b-intercept...
    Last edited by >_<SHY_GUY>_<; April 26th 2009 at 01:28 PM.
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