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Math Help - cosine derivative

  1. #1
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    cosine derivative

    Differentiate: y = \frac {1 + cosx} {1- cosx}
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Samantha View Post
    Differentiate: y = \frac {1 + cosx} {1- cosx}
    Use the quotient rule:

    <br />
f(x)=\frac{g(x)}{h(x)}<br />

    then:

    <br />
f'(x)=\frac{g'(x)h(x)-g(x)h'(x)}{(h(x))^2}<br />

    here g(x)=1 + \cos(x) and h(x)=1 - \cos(x), and so:

    <br />
g'(x)=-\sin(x)<br />

    and:

    <br />
h'(x)=\sin(x)<br />

    Putting this all together:

    <br />
f'(x)=\frac{-\sin(x)(1-\cos(x))-(1+\cos(x))\sin(x)}{(1-\cos(x))^2}<br />

    which you can simplify for yourself.

    RonL
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Samantha View Post
    Differentiate: y = \frac {1 + cosx} {1- cosx}
    y = \frac {1 + cos(x)} {1- cos(x)}

    Using the quotient rule:
    y^{\prime} = \frac{-sin(x) \cdot (1 - cos(x)) - (1 + cos(x)) \cdot sin(x)}{(1 - cos(x))^2}

    y^{\prime} = \frac{-sin(x) + sin(x) cos(x) - sin(x) - sin(x) cos(x)}{(1 - cos(x))^2}

    y^{\prime} = \frac{-2sin(x)}{(1 - cos(x))^2}

    We can play around with the denominator, but I see no value in that. (1 - cos(x))^2 = 1 - 2cos(x) + cos^2(x) which gives us no cancellations that I can see.

    -Dan
    Last edited by topsquark; July 9th 2007 at 08:48 AM.
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  4. #4
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    Find dy/dx for y= (x^3) {\sqrt x+1}
    Last edited by Samantha; July 9th 2007 at 06:06 AM.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Samantha View Post
    Find dy/dx for y= (x^3) \sqrt x+1
    Please use a new post for a new question.

    y= (x^3) \sqrt x+1

    The simple way:

    y= x^{7/2}+1

    So
    y^{\prime} = \frac{7}{2}x^{5/2}

    -Dan
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  6. #6
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    Quote Originally Posted by topsquark View Post
    Please use a new post for a new question.

    y= (x^3) \sqrt x+1

    The simple way:

    y= x^{7/2}+1

    So
    y^{\prime} = \frac{7}{2}x^{5/2}

    -Dan
    I'm not sure if I wrote it right. It's suppose to be square root of x+1
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  7. #7
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    y=x^3\sqrt{x+1}
    y'=(x^3)'\sqrt{x+1}+(x^3)\left(\sqrt{x+1}\right)'
    3x^2\sqrt{x+1}+x^3 \cdot \frac{1}{2}\cdot \frac{1}{\sqrt{x+1}}
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  8. #8
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    I didn't get it =(
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  9. #9
    Forum Admin topsquark's Avatar
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    ThePerfectHacker used the multiplication rule:
    (fg)^{\prime} = f^{\prime} \cdot g + f \cdot g^{\prime}

    So we want to find the derivative of x^3 \sqrt{x + 1}
    then f(x) = x^3 and g(x) = \sqrt{x + 1}.

    The derivative of \sqrt{x + 1} is \frac{1}{2\sqrt{x + 1}}.

    So
    (x^3 \sqrt{x + 1})^{\prime} = (x^3)^{\prime} \cdot \sqrt{x + 1} + x^3 \cdot (\sqrt{x + 1})^{\prime}

    (x^3 \sqrt{x + 1})^{\prime} = 3x^2 \sqrt{x + 1} + x^3 \cdot \frac{1}{2\sqrt{x + 1}}

    as TPH said.

    -Dan
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  10. #10
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    Hello, Samantha!

    If you want to surprise/impress/terrify your teacher, use these identities:

    . . \sin^2\!\left(\frac{\theta}{2}\right) \: = \:\frac{1 - \cos\theta}{2} \quad\Rightarrow\quad 1 - \cos\theta \:=\:2\sin^2\!\left(\frac{\theta}{2}\right)

    . . \cos^2\!\left(\frac{\theta}{2}\right) \: = \: \frac{1 + \cos\theta}{2}\quad\Rightarrow\quad1 + \cos\theta \:=\:2\cos^2\!\left(\frac{\theta}{2}\right)


    Differentiate: . y \:= \:\frac {1 + \cos x} {1- \cos x}

    We have: . y\;=\;\frac{1 + \cos x}{1 - \cos x} \;=\;\frac{2\cos^2\!\left(\frac{x}{2}\right)}{2\si  n^2\!\left(\frac{x}{2}\right)} \;= \;\cot^2\!\left(\frac{x}{2}\right)

    Then: . y' \;=\;2\cot\left(\frac{x}{2}\right)\cdot\left[-\csc^2\!\left(\frac{x}{2}\right)\right] \cdot\frac{1}{2}


    . . Therefore: . \boxed{\;y' \;= \;-\csc^2\!\left(\frac{x}{2}\right)\cot\left(\frac{x}  {2}\right)}

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  11. #11
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, Samantha!

    If you want to surprise/impress/terrify your teacher, use these identities:

    . . \sin^2\!\left(\frac{\theta}{2}\right) \: = \:\frac{1 - \cos\theta}{2} \quad\Rightarrow\quad 1 - \cos\theta \:=\:2\sin^2\!\left(\frac{\theta}{2}\right)

    . . \cos^2\!\left(\frac{\theta}{2}\right) \: = \: \frac{1 + \cos\theta}{2}\quad\Rightarrow\quad1 + \cos\theta \:=\:2\cos^2\!\left(\frac{\theta}{2}\right)



    We have: . y\;=\;\frac{1 + \cos x}{1 - \cos x} \;=\;\frac{2\cos^2\!\left(\frac{x}{2}\right)}{2\si  n^2\!\left(\frac{x}{2}\right)} \;= \;\cot^2\!\left(\frac{x}{2}\right)

    Then: . y' \;=\;2\cot\left(\frac{x}{2}\right)\cdot\left[-\csc^2\!\left(\frac{x}{2}\right)\right] \cdot\frac{1}{2}


    . . Therefore: . \boxed{\;y' \;= \;-\csc^2\!\left(\frac{x}{2}\right)\cot\left(\frac{x}  {2}\right)}

    (Chuckles) I like it.

    -Dan
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