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ANother C3 Question!! Urgent watch

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    (Original post by plastercaster)
    yes.
    PS. according to my source, his differentiation is wrong. at any rate, you can't differentiate inverse sine yet.
    no we haven't been taught how to differentiate inverse of sin,cos or tan.


    here is what i got for my answer

    dy/dx= 1/8cos(2sin^-1(x/4)+3)
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    (Original post by tgodkin)
    x=ln(2y+6)
    e^{x}=2y+6
    e^{x}-6=2y
    \frac{1}{2}e^{x}-3=y
    thanks man, but one thing is my differentiation correct?
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    (Original post by plastercaster)
    yes.
    PS. according to my source, his differentiation is wrong. at any rate, you can't differentiate inverse sine yet.
    Its a standard integral, but the proof is:

    Take y=sin^{-1}x

    sin(y)=x
    \frac{dx}{dy}=cos(y)
    \frac{dy}{dx}=\frac{1}{cos(y)}
    \frac{dy}{dx}=\frac{1}{\sqrt{1-sin^{2}(y)}}
    We know that sin(y)=x so:
    \frac{dy}{dx}=\frac{1}{\sqrt{1-x^{2}}}
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    the differential I posted is from wolfram alpha- http://www.wolframalpha.com/input/?i...n%28x%2F4%29-3
    It should be right.
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    they want it in term of x. so therefor you have to find cos(2y+6) in terms of x so root 1-(x^2/16)
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    (Original post by plastercaster)
    the differential I posted is from wolfram alpha- http://www.wolframalpha.com/input/?i...n%28x%2F4%29-3
    It should be right.
    You're correct, I forgot to square the \frac{1}{4}. Other than that the answers are the same.
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    Thanks to everyone for their input, I decided to just learn the rule of differentiating the inverse of sin and cos incase the question asks to give the answer in terms of x. I couldn't bothered with tan because is highly likely they gonna ask me to differentiate the inverse of tan.
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    (Original post by Remarqable M)
    Thanks to everyone for their input, I decided to just learn the rule of differentiating the inverse of sin and cos incase the question contains in terms of x. I couldn't bothered with tan because is highly likely they gonna ask me to differentiate the inverse of tan.
    \frac{d}{dx}tan^{-1}(x)=\frac{1}{1+x^2} :P

    Its simpler to prove than the other 2.
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    (Original post by Remarqable M)
    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).
    get it in terms of y= ....

    x/4=sin(2y+6)
    arcsin(x/4) = 2y+6
    arcsin(x/4) - 6 = 2y
    y= 1/2arcsin(x/4) - 3

    dy/dx = 1/(8arcsin(x/4))

    ive got it tomoro guessing you do too?..
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    (Original post by mn0710)
    get it in terms of y= ....

    x/4=sin(2y+6)
    arcsin(x/4) = 2y+6
    arcsin(x/4) - 6 = 2y
    y= 1/2arcsin(x/4) - 3

    dy/dx = 1/(8arcsin(x/4))

    ive got it tomoro guessing you do too?..
    if you mean the exam? yeah i got it tomorrow
    just revising how to solve equations now then im all done.
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    (Original post by Remarqable M)
    if you mean the exam? yeah i got it tomorrow
    just revising how to solve equations now then im all done.
    Was my explanation not good enough for you...? :p:
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    (Original post by Mathematician!)
    Was my explanation not good enough for you...? :p:
    HAHA it is now I just abandoned this thread lol and went back to the old thread your method is flawless and rather easy.:p:
    Im planning to double rep++

    If you don't mind can you have a look at my next question in this thread: http://www.thestudentroom.co.uk/show....php?t=1157616
 
 
 
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