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Math Help - differentiation quiery

  1. #1
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    differentiation quiery

    if im having to use the definition of the derivative \frac{(x+h)-f(x)}{h} to show if f(x)=sin2x then f'(x)=2cos2x how would i do this.i can do it for normal non trig functions?thanks sum difference formula?
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  2. #2
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    Quote Originally Posted by skystar View Post
    if im having to use the definition of the derivative \frac{(x+h)-f(x)}{h} to show if f(x)=sin2x then f'(x)=2cos2x how would i do this.i can do it for normal non trig functions?thanks


    You'll need to use the comount angle forum on \sin(2(x +h)) = \sin2x \cos 2h + \cos 2x \sin 2h

    and you'll need to use the following limits \lim_{X \rightarrow 0} \left( \frac{\sin X}{X} \right) = 1 and \lim_{X \rightarrow 0} \left( \frac{1-\cos X}{X} \right)  = 0

    Bobak
    Last edited by bobak; June 3rd 2008 at 03:28 AM. Reason: fixed an little error
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  3. #3
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    ive expanded it outand put over h, i get cos2x im a bit confused about the cos(h)/h part i think. thanks
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  4. #4
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    \frac{sin2x(cos2h-1)+cos2xsin2h}{h} thats where im at and now im just stuck,thanks,
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  5. #5
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    Quote Originally Posted by skystar View Post
    ive expanded it outand put over h, i get cos2x im a bit confused about the cos(h)/h part i think. thanks
    I made a little errror in my post above Its fixed now.


    f'(x) = \lim_{h \rightarrow 0} \left( \frac{\sin2x \cos 2h + \cos 2x \sin 2h-\sin 2x}{h} \right )<br />

    \Rightarrow \lim_{h \rightarrow 0} \left( \frac{\sin2x (\cos 2h -1) + \cos 2x \sin 2h}{h} \right )

    multiply the top and bottom by 2

    \Rightarrow \lim_{h \rightarrow 0} \left( \frac{2\sin2x (\cos 2h -1) + 2\cos 2x \sin 2h}{2h} \right )

    Then continue.

    Bobak
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  6. #6
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    after
    <br />
\Rightarrow \lim_{h \rightarrow 0} \left( \frac{2\sin2x (\cos 2h -1) + 2\cos 2x \sin 2h}{2h} \right )<br />

    im still confused i can make 2sin2x into things like 4sinxcosx but i cant see a way to simplify.
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  7. #7
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    Quote Originally Posted by skystar View Post
    after
    <br />
\Rightarrow \lim_{h \rightarrow 0} \left( \frac{2\sin2x (\cos 2h -1) + 2\cos 2x \sin 2h}{2h} \right )<br />

    im still confused i can make 2sin2x into things like 4sinxcosx but i cant see a way to simplify.
    No don't apply any double angle formula.

    take a minus out of the cosine limit part and split the sum into two limits.

    <br />
\Rightarrow \lim_{h \rightarrow 0} \left( \frac{-2\sin2x (1 -\cos 2h)}{2h}\right ) + \lim_{h \rightarrow 0} \left( \frac{2\cos 2x \sin 2h}{2h} \right )<br />

    take anything that is independent of h outside the limit.

    <br />
\Rightarrow -2\sin2x \lim_{h \rightarrow 0} \left( \frac{ (1 -\cos 2h)}{2h}\right ) +  2\cos 2x \lim_{h \rightarrow 0} \left( \frac{\sin 2h}{2h} \right )<br />

    Now just apply the two limits I gave you in my first post.

    Bobak
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