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    please help
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    y= 5/x

    y= 5x^-1

    Then differentiate that
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    (Original post by penelopecrux)
    please help
    If we know this rule


    What is another way  \dfrac{5} {x} can be written as?
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    (Original post by econwarwicker)
    x
    Posting full solutions are against forum rules
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    (Original post by econwarwicker)
    y= 5/x

    y= 5x^-1

    dy/dx= -5x^-2

    but wouldn't 5x^-1 be 1/5x?
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    (Original post by KaylaB)
    Posting full solutions are against forum rules
    oh wow really I'll edit it
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    (Original post by penelopecrux)
    but wouldn't 5x^-1 be 1/5x?
    No. And be careful not to write ambiguous expressions.
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    (Original post by econwarwicker)
    oh wow really I'll edit it
    That'd be ace thanks, it's just so people are actually learning as opposed to just copying answers
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    (Original post by penelopecrux)
    but wouldn't 5x^-1 be 1/5x?
    Would it help if we first think of  \dfrac{5} {x} as  5(\dfrac{1} {x}) ?
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    (Original post by KaylaB)
    That'd be ace thanks, it's just so people are actually learning as opposed to just copying answers
    Yes I get that. But if they wanted the answer for a question they can easily get that from a mark scheme. Having solutions with the final answer and showing how to get to the final step is what helps, not just a small tip. For simple ones like this then yeah thats fine.
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    (Original post by penelopecrux)
    but wouldn't 5x^-1 be 1/5x?
    The negative power is only applied to the  x so 5x^{-1} = \frac{5}{x} or     5 \times \frac{1}{x}

    If you wanted it to be \frac{1}{5x} then it'd need to be (5x)^{-1} as this then applies the power to everything inside the bracket, as opposed to just the x.
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    (Original post by econwarwicker)
    Yes I get that. But if they wanted the answer for a question they can easily get that from a mark scheme. Having solutions with the final answer and showing how to get to the final step is what helps, not just a small tip. For simple ones like this then yeah thats fine.
    do u do economics at warwick
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    (Original post by KaylaB)
    Would it help if we first think of  \dfrac{5} {x} as  5(\dfrac{1} {x}) ?

    yes it does help, thank you!
    also, another question please:
    say if i wanted to multiply x^8 by 6x^5, would it be 6x^13
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    (Original post by penelopecrux)
    yes it does help, thank you!
    also, another question please:
    say if i wanted to multiply x^8 by 6x^5, would it be 6x^13
    No problem! :hat2:

    & yes that's correct :yep:
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    (Original post by penelopecrux)
    also, another question please:
    say if i wanted to multiply x^8 by 6x^5, would it be 6x^13
    Yes.
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    after a lot of practice you will get it. Also, there are like `rules´which will help you learn how to calculate these
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    (Original post by penelopecrux)
    do u do economics at warwick
    no but im going to do it
 
 
 
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