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Thread: The derivative of sin(x)

  1. #1
    May 2011

    The derivative of sin(x)

    The difference quotient is:

    \frac{d}{dx}sin(x) = \lim \Delta x\rightarrow 0=\frac{sin(x+\Delta x)-sin(x)}{\Delta x}

    which converts to:

    =\lim \Delta x\rightarrow 0\frac{sin(x)cos\Delta x+cos(x)sin(\Delta x)-sin(x)}{\Delta x}

    The above I understand. The below step is supposed to be an algebraic rearrangement of the above:

    \frac{d}{dx}sin(x)=\lim \Delta x\rightarrow 0[cos(x)(\frac{sin(\Delta x)}{\Delta x})-sin(x)(\frac{1-cos(\Delta x)}{\Delta x})]

    I am wondering how things were changed to the step above. I am not seeing it right now.

    Thanks in advance...
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  2. #2
    MHF Contributor
    Nov 2010

    Re: The derivative of sin(x)

    \dfrac{ab+cd}{e} = a\dfrac{b}{e} + c\dfrac{d}{e} = \dfrac{ab}{e} + \dfrac{cd}{e} (this is order of operations and associativity of multiplication). Apply that to what you had. Consider the terms of the numerator. The first and last term have \sin(x). Factor it from those terms, and you get \sin(x)(\cos(\Delta x)-1). Put a negative sign in front, and you get -\sin(x)(1-\cos(\Delta x)). The middle term has a \cos(x).
    Thanks from sepoto
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