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    Given that x= 4sin(2y+6), find dy/dx in terms of x. please work it out for me simple and clear(step by step explaining how you got each equation).
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    dx/dy=8cos(2y+6)

    dy/dx=1/(dx/dy)
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    (Original post by gilbo)
    dx/dy=8cos(2y+6)

    dy/dx=1/(dx/dy)
    i got dah aswell
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    (Original post by gilbo)
    dx/dy=8cos(2y+6)

    dy/dx=1/(dx/dy)

    can you do the full working please? because simply 1/dy/dx is not enough
    it wants it interms of x.
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    1/8cos(2y+6) i think..

    dy/dx sinx = cosx

    dy/dx 4sin(2x+6) would give you 8sin(2x+6)

    therefore

    dx/dy 4sin(2y+6) = 8cos(2y+6)

    dy/dx= 1/dx/dy = 1/ 8cos (2y+6)
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    (Original post by sahil112)
    i got dah aswell
    is incomplete, the question wants in terms of x and i am stuck on here.
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    oh **** yeh mu bad i think you need to do a substitution. u = _______ blady blah
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    (Original post by zeb191)
    1/8cos(2y+6) i think..

    dy/dx sinx = cosx

    dy/dx 4sin(2x+6) would give you 8sin(2x+6)

    therefore

    dx/dy 4sin(2y+6) = 8cos(2y+6)

    dy/dx= 1/dx/dy = 1/ 8cos (2y+6)
    but the question is askin in terms of x(your working is correct but incomplete and im stuck on here)
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    (Original post by gilbo)
    oh **** yeh mu bad i think you need to do a substitution. u = _______ blady blah
    care to do it please i need to know how do get the final answer in clear step by step
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    He needs it in terns of x and ive bloody forgotten how to do it, im getting the poster dw.
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    (Original post by gilbo)
    He needs it in terns of x and ive bloody forgotten how to do it, im getting the poster dw.
    Urgently! please I'm going to bed soon and i want to get this out of my head lol
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    x= 4sin(2y+6)

    y=\frac{1}{2}sin^{-1}(\frac{x}{4})-3 By making y the subject.
    \frac{dy}{dx}=\frac{1}{8\sqrt{1-\frac{x^{2}}{16}}}
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    x=4sin(2y+6)
    so x/4=sin w if w=2y+6

    so dy/d= 1/8cos w
    we know cos^2w+sin^2w=1 so cos w=1- sqrt sin2w

    so cs w = 1- sqrt x2/4^2

    so dy/dx=1/8 sqrt 1- x^2/16

    so dy/dx = 1/2 sqrt 16-x^2

    hope that kinda made sense... sqrt = square root, and so answer = 1/ 2 times square root of 16- x^2
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    x=4sin(2y+6)
    I'm going to use the chain rule.
    u=2y+6
    du/dy=2

    x=4sinu
    dx/du=4cosu

    Dx/dy=du/dy*dx/du=2x4cosU=8cosU
    Dy/Dx is the inverse of this so 1/(8cosU)

    To find in terms of x, simply find U in terms of x. (there may be a prettier method but I think this works)
    x=4sinU
    x/4=sinu
    u=arcsin(x/4)
    so dy/dx in terms of x is
    1/(8cos(arcsin(x/4)))

    EDIT: answer according to wolfram
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    (Original post by tgodkin)
    x= 4sin(2y+6)

    y=\frac{1}{2}sin^{-1}(\frac{x}{4})-3 By making y the subject.
    \frac{dy}{dx}=\frac{1}{8\sqrt{1-x^{2}}}
    how did you make y the subject???

    how did you get y out of the bracket???:confused:

    edit: i know how you got x/4 but then how did you get y on its own...

    x/4=sin(2y+6)....
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    (Original post by Remarqable M)
    how did you make y the subject???

    how did you get y out of the bracket???:confused:
    x=4sin(2y+6)
    \frac{x}{4}=sin(2y+6)
    sin^{-1}(\frac{x}{4})=2y+6
    sin^{-1}(\frac{x}{4})-6=2y
    \frac{1}{2}sin^{-1}(\frac{x}{4})-3=y
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    (Original post by tgodkin)
    x=4sin(2y+6)
    \frac{x}{4}=sin(2y+6)
    sin^{-1}(\frac{x}{4})=2y+6
    sin^{-1}(\frac{x}{4})-6=2y
    \frac{1}{2}sin^{-1}(\frac{x}{4})-3=y
    THANK YOU SO MUCH! It makes sense now! REP+++++

    This thread is no longer needed lol
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    One more thing say we have ln instead of sin example,

    x=ln(2y+6) if i want y is it y=e^x-6/2 ??
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    yes.
    PS. according to my source, his differentiation is wrong. at any rate, you can't differentiate inverse sine yet.
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    x=ln(2y+6)
    e^{x}=2y+6
    e^{x}-6=2y
    \frac{1}{2}e^{x}-3=y
 
 
 
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