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C3 Question. Need help with collecting like terms to simplify Watch

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    It's page 136. Example 6 in the C3 Edexcel A-Level textbook.
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    There is more than one edexcel textbook, and there are many more exam boards than edexcel.
    The chances that someone will be online, on the same exam board as you, and have the same book are slim.
    post the question.
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    The thing is. It has like loads of powers and fractions and square roots etc. So it would look like a complete mess if I copied it down on this forum
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    See the sticky guide to posting on the maths forum.
    It should explain how to use LaTeX, which lets you
    raise^{things to powers} and \dfrac{do}{fractions}
    and stuff.

    either way, if you don't post the question, you won't get help.
    oh, and of course, to  \sqrt{root} stuff.
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    take a picture of it perhaps?
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    (Original post by nitez)
    It's page 136. Example 6 in the C3 Edexcel A-Level textbook.
    I've got the question, i think. f(x) = x^2 \sqrt(3x-1) and finding f'(x)?

    You've got:
    

f'(x) = x^2 \times \frac{3}{2} (3x-1)^{-\frac{1}{2}} + \sqrt(3x-1) \times 2x

    I think whats confused you is the negative power.
    \frac{3}{2} (3x-1)^{-\frac{1}{2}}
    can also be written as:
    \dfrac{3}{2\sqrt(3x-1)}

    go from there.
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    Guessing you mean the newest textbook and it's on the chain rule, here you go:

    Given that f(x) = x^2\sqrt{3x-1} find f'(x)

    The example then proceeds to differentiate using the chain rule to get:
    f'(x) = x^2 \times \frac{3}{2}(3x-1)^{-\frac{1}{2}} + \sqrt{3x-1} \times 2x

    = \dfrac{3x^2 + 12x^2 - 4x}{2\sqrt{3x-1}}



= \dfrac{15x^2 - 4x}{2\sqrt{3x-1}}



= \dfrac{x(15x - 4)}{2\sqrt{3x-1}}

    Guessing OP's problem is understanding what happened between the first line after differentiating and the line where the fraction comes into it. Edexcel C3 examples do tend to show random fragments of workings and I'm half asleep so I actually can't see what they've done myself, guessing they've just taken a factor of (3x-1)^-0.5 out but too tired to think.
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    Yep that is the correct question and I am indeed confused how they got from the first line to where the fraction came into it
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    (Original post by nitez)
    Yep that is the correct question and I am indeed confused how they got from the first line to where the fraction came into it
    If you're doing C3, you should be aware of this:
    

ax^{-b} = \dfrac{a}{x^b}
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    (Original post by F1Addict)
    If you're doing C3, you should be aware of this:
    

ax^{-b} = \dfrac{a}{x^b}
    Oh I get that part.... just dont see how they got the top of the fraction
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    (Original post by nitez)
    Oh I get that part.... just dont see how they got the top of the fraction
    Oh ok, hmm..

    

\dfrac{3x^2}{2\sqrt{3x-1}} + 2x(\sqrt{3x-1})



= \dfrac{3x^2 + 2x(\sqrt{3x-1} \times 2\sqrt{3x-1}}{2\sqrt{3x-1}}



= \dfrac{3x^2 + 2x(2(3x - 1))}{2\sqrt{3x-1}}



= \dfrac{3x^2 + 2x(6x - 2)}{2\sqrt{3x-1}}



= \dfrac{3x^2 + 12x^2 - 4x}{2\sqrt{3x-1}}

    That should help.
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    I was having the exact same problem! What a silly thing to miss- multiplying top and bottom! I feel so stupid. Thank you so so much F1Addict, literally been racking my brain for so long!
 
 
 
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