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Further Differentiation AQA core 2 help! watch

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    Hello I was wondering if you guys could quickly help me?

    a) y= 16x + x^-1, find the two value for x for which dy/dx = 0 ( 5 marks)
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    Differentiate y and equate to 0?

    Remember x^-2 = 1/x^2

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    (Original post by F1Addict)
    Differentiate y and equate to 0?

    Remember x^-2 = 1/x^2

    oh right I got it as 1/the square root of x. how do i get the two x values?
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    (Original post by raveen789)
    oh right I got it as 1/the square root of x. how do i get the two x values?
    How did you get 1/square root x?

    If y = 16x + x^-1

    then, dy/dx = 16 - x^-2

    The x^2 means you'll get two x-values as you get a ± when you square root something.
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    (Original post by F1Addict)
    How did you get 1/square root x?

    If y = 16x + x^-1

    then, dy/dx = 16 - x^-2

    The x^2 means you'll get two x-values as you get a ± when you square root something.
    basically i got 1/x^2 = 16

    I don't know what to do from there do i square both side?
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    (Original post by raveen789)
    basically i got 1/x^2 = 16

    I don't know what to do from there do i square both side?
    Kinda. Multiply both sides by x^2 and solve for x
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    (Original post by F1Addict)
    Kinda. Multiply both sides by x^2 and solve for x
    so if you times both sides of the equation by X^2 you would get 1 = 16x^2 ...
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    (Original post by F1Addict)
    Kinda. Multiply both sides by x^2 and solve for x
    He doesnt have to do that.


    x^-2 = 16

    take the -2 root of 16.
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    Ends up +/- 1/4 because you need the square root of 1/16.
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    (Original post by marek35)
    He doesnt have to do that.


    x^-2 = 16

    take the -2 root of 16.
    Basically I have that but in the form:

    1/X^2 = 16

    I am not sure what to do after this, i mean i could easily take your answers but it would be really helpful if you could tell me why i need to do the following

    Thanks guys for helping me
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    (Original post by raveen789)
    Basically I have that but in the form:

    1/X^2 = 16

    I am not sure what to do after this, i mean i could easily take your answers but it would be really helpful if you could tell me why i need to do the following

    Thanks guys for helping me
    My first time using LaTeX, please excuse any embarrassing mistakes!

    Also not sure if this is the best way of explaining, but hopefully it should do more good than harm

    x^{-2} = 16

    \frac{1}{x^2} = 16

    \frac{1}{x^2} = \frac{16}{1}

    \frac{x^2}{1} = \frac{1}{16}

    x^2 = \frac{1}{16}

    x = \pm \sqrt \frac{1}{16}

    x = \pm \frac{1}{\sqrt 16}

    x = \pm \frac{1}{4}
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    (Original post by raveen789)
    Basically I have that but in the form:

    1/X^2 = 16

    I am not sure what to do after this, i mean i could easily take your answers but it would be really helpful if you could tell me why i need to do the following

    Thanks guys for helping me
    Well you need to do this as you're looking for what x= . not what 1/x^2= to.
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    Thanks guys for the help I appreciate it, i get it now
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    (Original post by marek35)
    He doesnt have to do that.


    x^-2 = 16

    take the -2 root of 16.
    Yeah, thats also possible, but I always try to minimize using a calculator so thats the way I would've done it. I don't think 'multiply both sides by x^2', but thats what I do instincitively. There are a few of ways to do it, but as long as you end up with ±0.25 its fine.
 
 
 
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