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    The answer is 1/root(2y) but i cant do it.
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    Wth?
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    (Original post by Simplicity)
    Wth?
    ??
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    There are two ways to do this.

    1. Write \sqrt{2y} as \sqrt{2}\sqrt{y} = \sqrt{2} \times y^{1/2} and differentiate as usual

    2. Use the chain rule, with u = 2y

    Note that the answer to #1 will actually give you \dfrac{\sqrt{2}}{2\sqrt{y}}, but since \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}, it's equivalent. The answer to #2 gives it immediately in the form required.
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    Saying differentiate root 2y is ambigous.

    Anyway,

    \frac{d(\sqrt{2y})}{dy}=\frac{1}  {2} \times 2 \times \frac{1}{\sqrt{2y}}

    Do you know the chain rule?
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    (Original post by nuodai)
    There are two ways to do this.

    1. Write \sqrt{2y} as \sqrt{2}\sqrt{y} = \sqrt{2} \times y^{1/2} and differentiate as usual

    2. Use the chain rule, with u = 2y

    Note that the answer to #1 will actually give you \dfrac{\sqrt{2}}{2\sqrt{y}}, but since \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}, it's equivalent. The answer to #2 gives it immediately in the form required.
    it days that i dont need to use the product rule, quotient or chain rule if i do the algebra first.

    can you show me this method.
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    (Original post by Mr Kipling)
    it days that i dont need to use the product rule, quotient or chain rule if i do the algebra first.

    can you show me this method.
    Do you mean the whole \displaystyle \mbox{f}'(x) = \lim_{h \to 0}\dfrac{\mbox{f}(x+h) - \mbox{f}(x)}{h} thing? Or can you use the result \dfrac{\mbox{d}}{\mbox{d}x} \left( ax^n \right) = nax^{n-1}?
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    (Original post by Mr Kipling)
    it days that i dont need to use the product rule, quotient or chain rule if i do the algebra first.

    can you show me this method.
    I guess it would be this

    \frac{\sqrt{y+dy}- \sqrt {y}}{dy}
    then difference of squares.

    But, that isn't easier.
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    (Original post by Mr Kipling)
    it days that i dont need to use the product rule, quotient or chain rule if i do the algebra first.

    can you show me this method.
    That was nuodai's first method. (They were two separate methods.)
 
 
 
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