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Math Help - Help - Basic functions

  1. #1
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    Exclamation Help - Basic functions

    Can someone give me some guidance please, just started new calculas assignment and need some help:

    If f(theta) = 10Sin^2(0.4theta - 0.1)
    Show that f'(theta) = 4Sin(0.8theta - 0.2)

    and evaluate f''(pie/4) to 3 decimal places?

    For first part do I first change sin to cos and then remove the brackets which would give me 4Cos^2 but not much help with the rest?
    I should get further than this but just need some help getting started.

    Any help much appreciated.
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  2. #2
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    Hi

    f(\theta) = 10 \sin^2(0.4 \theta - 0.1)

    f(\theta) = 10 (\sin(0.4 \theta - 0.1))^2 = 10 \:u(\theta)^2 where u(\theta) = \sin(0.4 \theta - 0.1)

    The derivative of (u(\theta))^2 is 2\:u(\theta)\;u'(\theta)

    Therefore you have to find the derivative of u(\theta) = \sin(0.4 \theta - 0.1)

    The derivative of f(g(\theta)) is f'(g(\theta))\:g'(\theta)

    Can you go on now ?
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  3. #3
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    ??

    Thanks for reply,

    Sorry but you have lost me now. I might have missed something, why use u ang g? I thought to get the solution i needed to differentiate into dy/dx= and so find f' =??
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  4. #4
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    I am sorry if I lost you ...

    You are right : you need to differentiate f with respect to \theta in order to find f'(\theta)

    I used u, f and g to help but it seems that I missed the objective !
    BTW I should have used g and h since f is already taken
    u(\theta) = \sin(0.4 \theta-0.1) = g(h(\theta)) where g(\theta)=\sin \theta and h(\theta)=0.4 \theta-0.1
    u'(\theta) = g'(h(\theta)) \:h'(\theta)

    u, g and h are intermediate functions aimed at using the usual formulas, that's all
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  5. #5
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    Thanks

    Thanks,

    Makes it a bit more clearer, I can see where the 0.8 and 0.2 come from but how does the 10sin^2 become 4sin, I thought this was by multiplying out the brackets but I'm not sure now?
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  6. #6
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    The differentiation of u(\theta) = \sin(0.4 \theta - 0.1) is u'(\theta) = 0.4\:\cos(0.4 \theta - 0.1)

    Therefore the differentiation of (\sin(0.4 \theta - 0.1))^2 is
    2\:\sin(0.4 \theta - 0.1)\:\:0.4\:\cos(0.4 \theta - 0.1)=0.4\:\:2\:\sin(0.4 \theta - 0.1)\:\cos(0.4 \theta - 0.1)

    Using trig identity \sin(2x)=2\:\sin x\:\cos x

    The differentiation of (\sin(0.4 \theta - 0.1))^2 is 0.4\:\sin(0.8 \theta - 0.2)

    Now you just need to multiply this result by 10 to get the derivative of f

    I hope this makes sense
    Last edited by running-gag; March 14th 2009 at 03:50 AM. Reason: Brackets forgotten
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