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Math Help - Very annoying differentials

  1. #1
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    Very annoying differentials

    right ive been given some questions for tomorrow after a 5 month break from university and i have completely lost the technique

    1) using product rule find f'(t) of cos t sin t

    i manage to get (-sin t)^2 (cos t)^2


    2) use quotient rule to find f'(t) of cos t / sin t

    once again i manage to get - sin^2 t - cos^2 t / sin^2 t

    i cant help but think i should be getting very simple answers?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Revolt View Post
    right ive been given some questions for tomorrow after a 5 month break from university and i have completely lost the technique

    1) using product rule find f'(t) of cos t sin t

    i manage to get (-sin t)^2 (cos t)^2
    How do you get that:

    d/dt[cos(t)sin(t)] = -sin^2(t) + cos^2(t) = cos(2t)

    The product rule is:

    d/dx[g(x)h(x)]=g'(x)h(x)+g(x)h'(x)

    RonL
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  3. #3
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    Quote Originally Posted by Revolt View Post
    1) using product rule find f'(t) of cos t sin t

    i manage to get (-sin t)^2 (cos t)^2
    Another attempt

    f(t)=\cos t\sin t=\frac12\sin2t\implies f'(t)=\cos2t
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    How do you get that:

    d/dt[cos(t)sin(t)] = -sin^2(t) + cos^2(t) = cos(2t)

    The product rule is:

    d/dx[g(x)h(x)]=g'(x)h(x)+g(x)h'(x)

    RonL

    i get the bit in bold. is that just another way to write cos (2t) or is there another step involved, my identities are poor :/

    have cracked no1 but no2 does not seem to get any simpler :/
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Revolt View Post
    i get the bit in bold. is that just another way to write cos (2t) or is there another step involved, my identities are poor :/

    have cracked no1 but no2 does not seem to get any simpler :/
    It is just a bog standard trig identity

    RonL
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Revolt View Post
    2) use quotient rule to find f'(t) of cos t / sin t

    once again i manage to get - sin^2 t - cos^2 t / sin^2 t

    i cant help but think i should be getting very simple answers?
    First do you know the quotient rule?

    d/dt[g(t)/h(t)] = [g'(t)h(t) - g(t)h'(t)]/[h(t)]^2

    Here g(t)=cos(t), and h(t)=sin(t).

    RonL
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    First do you know the quotient rule?

    d/dt[g(t)/h(t)] = [g'(t)h(t) - g(t)h'(t)]/[h(t)]^2

    Here g(t)=cos(t), and h(t)=sin(t).

    RonL
    yeah, ive managed to get it down to -1 / sin^2 t

    think ill not bother with the rest of the questions lol, will just see the guy tomorrow thanks for the help
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