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Math Help - Differential Coefficient

  1. #1
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    Differential Coefficient

    Find the differential coefficient for the following different functions to demonstrate the use of function of a function; the product and quotient rules:

    a) Y=(1+2x) squared
    B) y = 6x squared Sin 2x
    c) s= 2x squared
    Cos3x


    Can you help differentiate it please, thanks.
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    Hi Zacky

    Quote Originally Posted by zacky08 View Post

    a) Y=(1+2x) squared


    Use the chain rule here

    \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}

    make u = 1+2x and y=u^{2}


    Quote Originally Posted by zacky08 View Post


    B) y = 6x squared Sin 2x


    Use the product rule here

    y = u\times v \Rightarrow y' = vu'+uv'

    make u = 6x^2 and v = \sin(2x)


    Quote Originally Posted by zacky08 View Post


    c) s= 2x squared
    Cos3x

    Use the quotient rule here

    For y = \frac{u}{v} \Rightarrow y' = \frac{vu'-uv'}{v^2}

    make u = 2x^2 and v=\cos(3x)
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  3. #3
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    Use the chain rule here

    make u = 1+2x and y=u^{2}

    Here's how to start....

    First from u = 1+2x find \frac{du}{dx}

    Spoiler:
    \frac{du}{dx} = 2


    then from y=u^{2} find \frac{dy}{du}

    Spoiler:
    \frac{dy}{du} = 2u = 2(1+2x)


    After you have these, put them into the RHS of the equation below.

    \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}
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  4. #4
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    Thank you so much, could you be so kind to do for the other 2 questions aswell if possible, thanks gracefully, Zacky.
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    Quote Originally Posted by zacky08 View Post
    Thank you so much, could you be so kind to do for the other 2 questions aswell if possible, thanks gracefully, Zacky.
    The idea in the following 2 questions are the same as the previous post. You need to post your attempts on this forum. It is Math 'Help' Forum not Math 'Do' Forum

    Also google the product and quotient rules. There are heaps of examples out there.

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  6. #6
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    Use the product rule here



    make and .

    Ok firstly, i do 6x + sin(2x) + sin(2x) + 6x which gives me sin12x + sin12x = 2sin24x4. Am i on the right lines?
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  7. #7
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    Quote Originally Posted by zacky08 View Post

    Ok firstly, i do 6x + sin(2x) + sin(2x) + 6x which gives me sin12x + sin12x = 2sin24x4. Am i on the right lines?
    No this is all types of wrong.

    Firstly there is a difference between u and u' . u' is the derivative of u

    For example if u = x^2 then u' = 2x

    Do you understand this concept?

    Also

     <br />
\sin(2x) + \sin(2x) =  2\sin(2x) \neq \sin(12x^2) <br /> <br />
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  8. #8
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    Ah ok, i kind of grasp it now.

    So, Sin (2x) + 12x then 12x + Sin (2x) then?

    Which would be Sin(24x) + Sin(24x) = 2Sin (48x)
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  9. #9
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    No, not even close.

    This is the rule.

    y = u\times v \Rightarrow y' = vu'+uv'

    You need to find u' and v' using the following information.

    u = 6x^2 and v = \sin(2x)

    I get u' = 12x and v' = 2\cos(2x)

    The real question is, do you know how I found these?
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  10. #10
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    6x differentiated is 12x, and sin is differentitated into cos and same thing with (2x) to give 2cos(2x). Yeah i realised what i did wrong, still a bit vague. But now that i got and , what would be the next step? do i put this into the Dy by Dx again?
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  11. #11
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    The answer is  y' = vu'+uv'

    So from your working

     y' = 12x\sin(2x)+12x^2\cos(2x)

    You should also factor out 12x giving

     y' = 12x(\sin(2x)+x\cos(2x))
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  12. #12
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    Finally, the Quotient Rule.




    make and

    I would get.

    Cos (12x) - Cos (12x)
    2cos(9x)
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  13. #13
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    Quote Originally Posted by zacky08 View Post
    Finally, the Quotient Rule.




    make and

    I would get.

    Cos (12x) - Cos (12x)
    2cos(9x)
    One step at a time there big fella, firstly what did you get for u' and v' ?
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  14. #14
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    i got 4x and i got -sin (9x)
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  15. #15
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    u' is correct v' = -3\sin(3x)
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