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Math Help - derivative question, please help!

  1. #1
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    derivative question, please help!

    I don't know how to take out Δx when i do the question

    If y=sin x, show that

    <br />
\frac{dy}{dx}= cos x<br />

    (by the first principles)
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ling_c_0202
    I don't know how to take out Δx when i do the question

    If y=sin x, show that

    <br />
\frac{dy}{dx}= cos x<br />

    (by the first principles)
    \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{sin(x + \Delta x) - sin(x)}{\Delta x}

    Now, sin(A + B) = sin(A)cos(B) + sin(B)cos(A):

    = \lim_{\Delta x \to 0} \frac{sin(x)cos( \Delta x) + sin( \Delta x)cos(x) - sin(x)}{\Delta x}

    = \lim_{\Delta x \to 0} \left ( \frac{sin(x)cos( \Delta x)}{ \Delta x} + \frac{sin( \Delta x)cos(x)}{ \Delta x} - \frac{sin(x)}{\Delta x} \right )

    = \lim_{\Delta x \to 0} \left ( \frac{sin(x)cos( \Delta x)}{ \Delta x} \right ) + \lim_{\Delta x \to 0} \left ( \frac{sin( \Delta x)cos(x)}{ \Delta x} \right ) - \lim_{\Delta x \to 0} \left ( \frac{sin(x)}{\Delta x} \right )

    Now, \lim_{\Delta x \to 0} \frac{sin( \Delta x)}{ \Delta x} = 1 and \lim_{\Delta x \to 0} cos( \Delta x) = 1 so:

    \frac{dy}{dx} = \lim_{\Delta x \to 0} \left ( \frac{sin(x)}{ \Delta x} \right ) + cos(x) - \lim_{\Delta x \to 0} \left ( \frac{sin(x)}{\Delta x} \right )

     = cos(x).

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark
    \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{sin(x + \Delta x) - sin(x)}{\Delta x}

    Now, sin(A + B) = sin(A)cos(B) + sin(B)cos(A):

    = \lim_{\Delta x \to 0} \frac{sin(x)cos( \Delta x) + sin( \Delta x)cos(x) - sin(x)}{\Delta x}

    = \lim_{\Delta x \to 0} \left ( \frac{sin(x)cos( \Delta x)}{ \Delta x} + \frac{sin( \Delta x)cos(x)}{ \Delta x} - \frac{sin(x)}{\Delta x} \right )

    = \lim_{\Delta x \to 0} \left ( \frac{sin(x)cos( \Delta x)}{ \Delta x} \right ) + \lim_{\Delta x \to 0} \left ( \frac{sin( \Delta x)cos(x)}{ \Delta x} \right ) - \lim_{\Delta x \to 0} \left ( \frac{sin(x)}{\Delta x} \right )
    You can't do it this way as you are going to have to subtract infnities
    to get the finite limit - maybe OK in Physics but not here

    Now, \lim_{\Delta x \to 0} \frac{sin( \Delta x)}{ \Delta x} = 1 and \lim_{\Delta x \to 0} cos( \Delta x) = 1 so:

    \frac{dy}{dx} = \lim_{\Delta x \to 0} \left ( \frac{sin(x)}{ \Delta x} \right ) + cos(x) - \lim_{\Delta x \to 0} \left ( \frac{sin(x)}{\Delta x} \right )

     = cos(x).

    -Dan
    RonL
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack
    You can't do it this way as you are going to have to subtract infnities
    to get the finite limit - maybe OK in Physics but not here



    RonL
    Doh! Sorry!

    -Dan
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  5. #5
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    Let die Meister try.
    ---
    You need to find, for the point x.
    \lim_{\Delta x\to 0}\frac{\sin(x+\Delta x)-\sin x}{\Delta x}
    \lim_{\Delta x\to 0}\frac{\sin x\cos \Delta x+\cos x \sin \Delta x-\sin x}{\Delta x}
    \lim_{\Delta x\to 0}\left[\sin x\left( \frac{\cos \Delta x -1}{\Delta x}\right)+\cos x\left( \frac{\sin \Delta x}{\Delta x} \right) \right]
    Recognize the famous limits,
    \lim_{\Delta x\to 0}\frac{\sin \Delta x}{\Delta x}=1
    \lim_{\Delta x\to 0}\frac{\cos \Delta x-1}{\Delta x}=0
    Since,
    \lim_{\Delta x\to 0}\frac{\cos \Delta x-1}{\Delta x}=0
    \lim_{\Delta x\to 0}\sin x=\sin x (limit of constant function).
    Therefore,
    \lim_{\Delta x\to 0}\sin x\left( \frac{\cos \Delta x -1}{\Delta x}\right)=0
    Because, "if two limits of two functions exist at a point then the limit of their product exists at the point and is equal to the product of their limits".
    Similary, since,
    \lim_{\Delta x\to 0}\frac{\sin \Delta x}{\Delta x}=1
    \lim_{\Delta x\to 0}\cos x=\cos x (limit of constant function).
    Therefore,
    \lim_{\Delta x\to 0}\cos x\left( \frac{\sin \Delta x}{\Delta x} \right)=\cos x
    Because, "if two limits of two functions exist at a point then the limit of their products exists at the point and is equal to the product of their limits"
    Thus,
    \lim_{\Delta x\to 0}\left[\sin x\left( \frac{\cos \Delta x-1}{\Delta x}\right)+\cos x\left( \frac{\sin \Delta x}{\Delta x}\right) \right]= 0+\cos x=\cos x
    Because, "if the limits of two functions exist at a point then the limit of the sum of these functions exists at the point and is equal to the sum of the limits".
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  6. #6
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    thank you

    Oh... I 've got it!! Thanks a lot!
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