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Thread: The derivative of sin(x)

  1. #1
    May 2011

    The derivative of sin(x)

    The difference quotient is:

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

    which converts to:

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

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

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

    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)

    $\displaystyle \dfrac{ab+cd}{e} = a\dfrac{b}{e} + c\dfrac{d}{e} = \dfrac{ab}{e} + \dfrac{cd}{e}$ (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 $\displaystyle \sin(x)$. Factor it from those terms, and you get $\displaystyle \sin(x)(\cos(\Delta x)-1)$. Put a negative sign in front, and you get $\displaystyle -\sin(x)(1-\cos(\Delta x))$. The middle term has a $\displaystyle \cos(x)$.
    Thanks from sepoto
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