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    Need help with this question:

    Given that x= 3tan(2y-1), find dy/dx in terms of x

    Thanks, +rep will be given tomorrow
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    3tan(2y-1)

    = 3 x 2sec^2(2y-1) = 6sec^2(2y-1)
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    yh i got 1/6sec^2(2y-1)
    because they want dy/dx not dx/dy

    but the answer at the back of the book is 3/2(9+x^2)

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    Hrmmm, your book is wack. IDK why they did that. But then again, I'm pretty slooow.
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    (Original post by Get Cape.Wear Cape.Fly.)
    Hrmmm, your book is wack. IDK why they did that. But then again, I'm pretty slooow.
    Lol its the official Edexcel C3 book..
    :o:
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    rearrange the original equation to make y the subject, then insert that into dy/dx and simplify (i haven't actually done it yet, but that should work)
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    (Original post by ScaryKid_ScaringKids)
    Lol its the official Edexcel C3 book..
    :o:
    dy/dx=1/6sec^2x (2y-1)

    You know x=3tan(2y-1)
    2y-1=arctan(x/3)
    sub it in

    1/6sec^2x.arctan(x/3)

    and then i'm screwed lol....not sure where to go from here but i hope that helps (if that is right)
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    Yh i tried doing that...
    but i got y=(tan^-1(x/3)+1)/2
    can i substitute that in dy/dx ?
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    (Original post by ScaryKid_ScaringKids)
    Yh i tried doing that...
    but i got y=(tan^-1(x/3)+1)/2
    can i substitute that in dy/dx ?
    (Original post by Superman_Jr)
    dy/dx=1/6sec^2x (2y-1)

    You know x=3tan(2y-1)
    2y-1=arctan(x/3)
    sub it in

    1/6sec^2x.arctan(x/3)

    and then i'm screwed lol....not sure where to go from here but i hope that helps (if that is right)
    To do this question fairly quickly, you can use implicit differentiation and need to know how to differentiate inverse trig functions. It's possible to do it using purely C3 knowledge though because you've all got up to

    1/6sec^2(2y-1) as well.

    Using implicit differentiate just because it's neater, the solution goes:

    x = 3tan(2y - 1)

    1 = 6sec^2(2y-1) dy/dx

    dy/dx = 1/(6sec^2(2y-1))

    Using trig identity 1 + tan^2x = sec^2x

    dy/dx = 1/(6 + 6tan^2(2y-1))

    we know that x^2 = 9tan^2(2y - 1)

    dy/dx = 1/(6 + (2/3)x^2)

    dy/dx = 1/((18 + 2x^2)/3)

    dy/dx = 3/18 + 2x^2

    dy/dx = 3/2(9 +x^2)
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    (Original post by ScaryKid_ScaringKids)
    Yh i tried doing that...
    but i got y=(tan^-1(x/3)+1)/2
    can i substitute that in dy/dx ?
    Why are you substituting? You got dx/dy right? Then rearrange the sec^2 bit into tan so it looks the same as the original equation (x=tan(2y-1)) and then rearrange that original equation to get tan(2y-1) equals x/3. "tan(2y-1) is in your differential equation thingy so you sub it in and everything is in terms of x.
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    (Original post by Clarity Incognito)
    To do this question fairly quickly, you can use implicit differentiation and need to know how to differentiate inverse trig functions. It's possible to do it using purely C3 knowledge though because you've all got up to

    1/6sec^2(2y-1) as well.

    Using implicit differentiate just because it's neater, the solution goes:

    x = 3tan(2y - 1)

    1 = 6sec^2(2y-1) dy/dx

    dy/dx = 1/(6sec^2(2y-1))

    Using trig identity 1 + tan^2x = sec^2x

    dy/dx = 1/(6 + 6tan^2(2y-1))

    we know that x^2 = 9tan^2(2y - 1)

    dy/dx = 1/(6 + (2/3)x^2)

    dy/dx = 1/((18 + 2x^2)/3)

    dy/dx = 3/18 + 2x^2

    dy/dx = 3/2(9 +x^2)
    you say that, but im fairly sure it's not on all C3 syllabuses (ie different boards) because I didn't know that. :P
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    You could differentiate equation as it stands wrt x

    x = 3tan(2y-1)

    1 = 3 2sec^2(2y-1) dy/dx chain rule twice

    1 = 6( 1 + tan^2(2y-1) ) dy/dx trig identity

    1 = 6( 1 + (x^2)/9 ) dy/dx substituting from original equation

    dy/dx = 1/(6 (1 + (x^2)/9))) dividing both sides and swapping round

    dy/dx = 3/2(9 + x^2) tidying up expression
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    (Original post by Toneh)
    you say that, but im fairly sure it's not on all C3 syllabuses (ie different boards) because I didn't know that. :P
    I know that the knowledge is definitely true of edexcel but surely knowing that if you divide sin^2x + cos^2x = 1 by cos^x, you get tan^2x + 1 = sec^2x will be knowledge of trig across the all the examining boards by C3.
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    (Original post by Clarity Incognito)
    I know that the knowledge is definitely true of edexcel but surely knowing that if you divide sin^2x + cos^2x = 1 by cos^x, you get tan^2x + 1 = sec^2x will be knowledge of trig across the all the examining boards by C3.
    fair enough
    didn't think of that
    fortunately we have very little trig in OCR MEI
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    dx/dy= 2sec^2 (2y-1)

    dy/dx=1/2sec^2 (2y-1)
    =1/2( tan^2 (2y-1) +1)
    =1/2(1+x^2)
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    dx/dy = 6sec^2 (2y-1)
    dy/dx= 1/ (6sec^2(2y-1))
    using tan^2x+1=sec^2x we get:
    dy/dx = 1/(6(1+tan^2(2y-1))
    dy/dx = 1/(6+(2/3)x^2)
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    how come we have different answers?
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    x = 3tan(2y - 1)

    2y = arctan(\frac {x}{3}) + 1

    2\frac {dy}{dx} = \frac {1}{3(1 + (\frac {x}{3})^2)}

    2\frac {dy}{dx} = \frac {1}{3(1 + \frac {x^2}{9})}

    2\frac {dy}{dx} = \frac {1}{3(\frac {9 + x^2}{9})}

    2\frac {dy}{dx} = \frac {9}{27 + 3x^2}

    \frac {dy}{dx} = \frac {9}{54 + 6x^2} = \frac {3}{18 + 2x^2} = \frac {3}{2(9 + x^2)}

    :awesome:
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    (Original post by Pheylan)
    x = 3tan(2y - 1)

    2y = arctan(\frac {x}{3}) + 1

    2\frac {dy}{dx} = \frac {1}{3(1 + (\frac {x}{3})^2)}

    2\frac {dy}{dx} = \frac {1}{3(1 + \frac {x^2}{9})}

    2\frac {dy}{dx} = \frac {1}{3(\frac {9 + x^2}{9})}

    2\frac {dy}{dx} = \frac {9}{27 + 3x^2}

    \frac {dy}{dx} = \frac {9}{54 + 6x^2} = \frac {3}{18 + 2x^2} = \frac {3}{2(9 + x^2)}

    :awesome:
    Not everyone knows the differentials of inverse trig functions, it certainly isn't C3 knowledge.
 
 
 
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