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    (Original post by the_13th)
    IB SL
    That explains why it's so complicated then. I'm still trying to give it a go but not really getting anywhere. Good luck figuring it out though =)
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    I don't know if this is posting the image or not, but simplify that and you should get your answer.. I think xP

    http://postimage.org/image/156b2vyg4
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    (Original post by the_13th)
    can you upload a picture of your paper with the solution? I'm stuck.
    Here it is in latex:

    

y = \sqrt \dfrac{x-3}{2x+5}



y = (\dfrac{x-3}{2x+5})^\frac{1}{2}



\dfrac{dy}{dx} = \dfrac{1}{2} \times (\dfrac{x-3}{2x+5})^\frac{-1}{2} \times \dfrac{11}{(2x+5)^2}





= \dfrac{1}{2} \times \dfrac{1}{\sqrt \frac{x-3}{2x+5}} \times \dfrac{11}{(2x+5)^2}





= \dfrac{11}{2(2x+5)^2 \sqrt \frac{x-3}{2x+5}}





= \dfrac{11}{2(2x+5)^\frac{3}{2} \sqrt{x-3}}





= \dfrac{11}{2\sqrt{x-3}\sqrt{(2x+5)^3}}

    The \dfrac{11}{(2x+5)^2} comes from using the quotient rule on \dfrac{x-3}{2x+5}

    I might've over complicated it, but hey, it does work.
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    (Original post by F1Addict)
    Here it is in latex:

    

y = \sqrt \dfrac{x-3}{2x+5}



y = (\dfrac{x-3}{2x+5})^\frac{1}{2}



\dfrac{dy}{dx} = \dfrac{1}{2} \times (\dfrac{x-3}{2x+5})^\frac{-1}{2} \times \dfrac{11}{(2x+5)^2}





= \dfrac{1}{2} \times \dfrac{1}{\sqrt \frac{x-3}{2x+5}} \times \dfrac{11}{(2x+5)^2}





= \dfrac{11}{2(2x+5)^2 \sqrt \frac{x-3}{2x+5}}





= \dfrac{11}{2(2x+5)^\frac{3}{2} \sqrt{x-3}}





= \dfrac{11}{2\sqrt{x-3}\sqrt{(2x+5)^3}}

    The \dfrac{11}{(2x+5)^2} comes from using the quotient rule on \dfrac{x-3}{2x+5}

    I might've over complicated it, but hey, it does work.
    I managed to get to line 3 but didn't know how to rearrange it! I see it now though.
    Can you explain how you got from line 5 to 6?
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    (Original post by laughylolly)
    I managed to get to line 3 but didn't know how to rearrange it! I see it now though.
    Can you explain how you got from line 5 to 6?
    \sqrt {\dfrac{x-3}{2x+5}} = \dfrac{\sqrt{x-3}}{\sqrt{2x+5}}

    Then you get (2x+5)^2 * (2x+5)^{-0.5}, so the power becomes 3/2.
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    (Original post by F1Addict)
    \sqrt {\dfrac{x-3}{2x+5}} = \dfrac{\sqrt{x-3}}{\sqrt{2x+5}}

    Then you get (2x+5)^2 * (2x+5)^{-0.5}, so the power becomes 3/2.
    Thank you, I actually understand it now! Do quotient questions at A2 sorta level get much harder than that?
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    Dont bother with the quotient rule; just use the product rule with a power of -1, it is far easier to remember.

    Let y=u(x)v(x), then dy/dx = v(du/dx) + u(dv/dx)
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    (Original post by laughylolly)
    Thank you, I actually understand it now! Do quotient questions at A2 sorta level get much harder than that?
    No problem

    I guess this question would be pretty tricky in an exam and you'd find it quite late into the paper. It could be made harder if you had the work out the actual value of the gradient a line at a given point, or something to do with turning points.
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    (Original post by F1Addict)
    Here it is in latex:

    

y = \sqrt \dfrac{x-3}{2x+5}



y = (\dfrac{x-3}{2x+5})^\frac{1}{2}



\dfrac{dy}{dx} = \dfrac{1}{2} \times (\dfrac{x-3}{2x+5})^\frac{-1}{2} \times \dfrac{11}{(2x+5)^2}





= \dfrac{1}{2} \times \dfrac{1}{\sqrt \frac{x-3}{2x+5}} \times \dfrac{11}{(2x+5)^2}





= \dfrac{11}{2(2x+5)^2 \sqrt \frac{x-3}{2x+5}}





= \dfrac{11}{2(2x+5)^\frac{3}{2} \sqrt{x-3}}





= \dfrac{11}{2\sqrt{x-3}\sqrt{(2x+5)^3}}

    The \dfrac{11}{(2x+5)^2} comes from using the quotient rule on \dfrac{x-3}{2x+5}

    I might've over complicated it, but hey, it does work.
    You seem to have got the right answer and used the right method, but I still don't understand it, really.

    To begin with your value for v^2 = (2x+5)^2

    But you are given denominator (which is v) as Square root of (2x+5) surely v^2 would be just equal to (2x+5), why do you get (2x+5)^2?
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    i gave up on quotient rule years ago. just use product rule for everything
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    (Original post by the_13th)
    You seem to have got the right answer and used the right method, but I still don't understand it, really.

    To begin with your value for v^2 = (2x+5)^2

    But you are given denominator (which is v) as Square root of (2x+5) surely v^2 would be just equal to (2x+5), why do you get (2x+5)^2?
    Because he's used the chain rule; he's differentiating the square root with respect to (x-3)/(2x+5) then multiplying by the derivative of (x-3)/(2x+5), which then has (2x+5)^2 on the bottom.
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    (Original post by the_13th)
    You seem to have got the right answer and used the right method, but I still don't understand it, really.

    To begin with your value for v^2 = (2x+5)^2

    But you are given denominator (which is v) as Square root of (2x+5) surely v^2 would be just equal to (2x+5), why do you get (2x+5)^2?
    I get what you mean, which is right, but I did it differently.

    Initally I let u = \dfrac{x-3}{2x+5}, and used chain rule. This means I was able to use the quotient rule without the square root sign, which makes things much simpler, for me anyways.

    Haddock got it right.
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    (Original post by F1Addict)
    Haddock got it right.
    Yay.
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    (Original post by F1Addict)
    No problem

    I guess this question would be pretty tricky in an exam and you'd find it quite late into the paper. It could be made harder if you had the work out the actual value of the gradient a line at a given point, or something to do with turning points.
    Ok. I'm doing Higher Maths at the moment (As level equivalent) and Advanced higher (A2 level) next year so I'll probably be taught all of this eventually...
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    (Original post by F1Addict)
    I get what you mean, which is right, but I did it differently.

    Initally I let u = \dfrac{x-3}{2x+5}, and used chain rule. This means I was able to use the quotient rule without the square root sign, which makes things much simpler, for me anyways.

    Haddock got it right.
    YEEEEEES!

    Now I can go and sleep in peace...

    Thanks for your help and explanation. Can't believe our teacher failed to teach us this method.

    Great job. Much obliged.
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    (Original post by the_13th)
    YEEEEEES!

    Now I can go and sleep in peace...

    Thanks for your help and explanation. Can't believe our teacher failed to teach us this method.

    Great job. Much obliged.
    Glad you managed to get it out.
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    (Original post by the_13th)
    You seem to have got the right answer and used the right method, but I still don't understand it, really.

    To begin with your value for v^2 = (2x+5)^2

    But you are given denominator (which is v) as Square root of (2x+5) surely v^2 would be just equal to (2x+5), why do you get (2x+5)^2?
    Here is how you can use the product rule to get the answer (yes, its quite hard to read - my apologies!)

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    (Original post by Jack-)
    Here is how you can use the product rule to get the answer (yes, its quite hard to read - my apologies!)

    thanks a lot
 
 
 
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