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    Given that x=4sin(2y+6) find dy/dx in terms of X

    What i did was found dx/dy to get 8cos(2y+6).. then found dy/dx as 1/8cos(2y+6)... however im not sure where to go from here. The mark scheme gets arcsins from somewhere>?:confused:

    help plz
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    Yep, so far; so good! But it wants the answer in terms of x; your answer is in terms of y. Now, using the original equation (x=4sin(2y+6)) get 2y+6 on its own. That'll get you your arcsins!
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    (Original post by jakezg)
    Given that x=4sin(2y+6) find dy/dx in terms of X

    What i did was found dx/dy to get 8cos(2y+6).. then found dy/dx as 1/8cos(2y+6)... however im not sure where to go from here. The mark scheme gets arcsins from somewhere>?:confused:

    help plz
    Hello,

    I would try taking the inverse of both sides.

    arcsin (x/4) = 2y + 6

    then differentiate. The formula for arcsin is in the formulae book for edexcel (FP2).
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    (Original post by jakezg)
    Given that x=4sin(2y+6) find dy/dx in terms of X

    What i did was found dx/dy to get 8cos(2y+6).. then found dy/dx as 1/8cos(2y+6)... however im not sure where to go from here. The mark scheme gets arcsins from somewhere>?:confused:

    help plz

    x=4sin(2y+6)

    dx/dy = 8cos(2y+6)

    Therefore dy/dx = 1/dx/dy = 1/8cos(2y+6)

    Using: Sin^2(2y+6)+cos^2(2y+6) = 1

    cos^2(2y+6) = 1 - Sin^2(2y+6)

    We know that x=4Sin(2y+6), therefore x/4 = Sin(2y+6) and so
    (x/4)^2 = Sin^2(2y+6)


    Hence, dy/dx = 1/8Root1-(x/4)^2
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    What's arcsines :eek: :eek: :eek: C3??
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    its the inverse sine function. you know, sin^(-1)x
 
 
 

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