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Math Help - how do you find a dervative

  1. #1
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    how do you find a dervative

    i was just wondering if anyone could help me find the derivative of a problem heres the problem:

    \lim_{h \to 0}  \frac{\sin (\pi+h)-\sin (\pi)}{h}

    i didn't write it right i mean lim below lim there should be h arrow 0 sin (pi+h)-sin pi underline over h
    how do you find the derivative of that problem and what rule do you use to solve the problem
    Last edited by CaptainBlack; March 23rd 2008 at 12:16 AM.
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  2. #2
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    Quote Originally Posted by janemba View Post
    i was just wondering if anyone could help me find the derivative of a problem heres the problem lim  sin (pi+h)-sin pi[/U]<br />
                                                                                                                                             h->h           h

    i didn't write it right i mean lim below lim there should be h arrow 0 sin (pi+h)-sin pi underline over h
    how do you find the derivative of that problem and what rule do you use to solve the problem
    1. Do you know the compound angle formula for sin(A + B)?
    2. Do you know the exact value of cos(pi) and sin(pi)?
    3. Do you know that \lim_{h \rightarrow 0} \frac{\sin h}{h} = 1?
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  3. #3
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    where you put sin h =1 i mean sin (pi+h)-sin pi
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  4. #4
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    Quote Originally Posted by janemba View Post
    where you put sin h =1 i mean sin (pi+h)-sin pi
    He means use the sum formula for \sin on \sin(\pi+h) then use the known values
    of \sin(\pi) and \cos(\pi) and you will be left with something that to eveluate you will need
    to know:

    \lim_{h\to 0}\frac{\sin(h)}{h}=1

    RonL
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  5. #5
    Member javax's Avatar
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    If you didn't solve yet, just like these guys told, I did like this and I think it's ok:

    sin(x+y) = sinxcosy + cosxsiny
    \lim_{x\to 0}{\frac{sinx}{x}} = 1

    then

    \lim_{h\to 0}{\frac{sin(\pi+h)-sin\pi}{h}} = \lim_{h\to 0}{\frac{sin\pi cosh+cos\pi sinh - sin\pi}{h}}
    = \lim_{h\to o}{\frac{sin\pi(cosh-1)+cos\pi sinh}{h}} = \lim_{h\to 0}{\frac{cos\pi sinh}{h}} = cos\pi \lim_{h\to 0}{\frac{sinh}{h}} = cos\pi = -1
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  6. #6
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    ok but how did you get that answer
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by javax View Post
    If you didn't solve yet, just like these guys told, I did like this and I think it's ok:

    sin(x+y) = sinxcosy + cosxsiny
    \lim_{x\to 0}{\frac{sinx}{x}} = 1

    then

    \lim_{h\to 0}{\frac{sin(\pi+h)-sin\pi}{h}} = \lim_{h\to 0}{\frac{sin\pi cosh+cos\pi sinh - sin\pi}{h}}
    = {\color{red} \lim_{h\to o}{\frac{sin\pi(cosh-1)+cos\pi sinh}{h}}} = \lim_{h\to 0}{\frac{cos\pi sinh}{h}} = cos\pi \lim_{h\to 0}{\frac{sinh}{h}} = cos\pi = -1
    the red line is unnecessary. we know that \sin \pi = 0, so we could just eliminate the first and last terms, and go to your next line
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  8. #8
    Member javax's Avatar
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    Quote Originally Posted by Jhevon View Post
    the red line is unnecessary. we know that \sin \pi = 0, so we could just eliminate the first and last terms, and go to your next line
    sure Jhevon, I just wanted to be like more 'clear'...:P

    Well janemba...I don't see anything else te make clear! Maybe you're not having much knowledge about trigonometry.
    Visit the trigonometry section...find some of basic trigonometric identities and try to understand the Trigonometric Circle.
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  9. #9
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    Quote Originally Posted by janemba View Post
    ok but how did you get that answer
    Perhaps if you gave simple yes or no answers to the three questions I asked in my very first post it would be easier to help you.

    On the face of it, it looks like you need to go back and thoroughly revise some of the more basic material before attempting this sort of question.
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