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    If A = y + 1/(y+3), find dA/dy, simplifying the answer.

    i got 1-(y+3)

    the answer in the back of the book is completely different, its:

    (y+2)(y+4) / (y+3)^2


    :confused:


    and also:

    x= 1/4y^2

    i got:

    x = 4y^-2
    = -8y^-3
    but the answers say:
    -2y^3
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    The book is right.

    You should have 1-1/(y+3)^2
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    (Original post by steve10)
    The book is right.

    You should have 1-1/(y+3)^2
    how?
    and you answer isn't what the book says.
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    (Original post by SeekerOfKnowledge)
    If A = y + 1/(y+3), find dA/dy, simplifying the answer.

    i got 1-(y+3)

    the answer in the back of the book is completely different, its:

    (y+2)(y+4) / (y+3)^2


    :confused:

    I dunno where you got that from...
    This is done using the quotient rule:
    \frac{dy}{dx}= \frac{v\times \frac{du}{dx} - u\times\frac{dv}{dx}}{v^2}

    so u= y+1
    v= y+3

    can you go on from here...when I did this I got something different to what the book says too:

    \frac{dA}{dy} = \frac{2}{(y+3)^2}...not sure if its right ~ I'll check it again.


    Edit: ohh wait did I miss read the question..
    did you mean
    y+\frac{1}{y+3} in this case you'd apply the quotient rule to the second part and this could give you the answer which the book has..
    • PS Helper
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    is it y + 1/(y+3) or (y +1)/(y+3), ????
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    its differentiation ques. not simplifying ,, use quotient rule in chapter 8
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    OP is either reading the wrong questions/ answers, is drunk or is just trolling!!!
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    (Original post by SeekerOfKnowledge)
    how?
    and you answer isn't what the book says.

    dA/dy = 1-1/(y+3)^2

    simplify the rhs and you get the book's answer
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    \frac{dA}{dy}=1+\frac{0(y+3)-1(1)}{(y+3)^2}

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

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

    = \frac{y^2+6y+8}{(y+3)^2}

    = \frac{(y+2)(y+4)}{(y+3)^2}



    EDIT: I'm an idiot, the chain rule will do:

    A=y+\frac{1}{y+3}=y+(y+3)^{-1}

    \frac{dA}{dy}=1-(y+3)^{-2}

    \frac{dA}{dy}=1-\frac{1}{(y+3)^2}
 
 
 
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