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    when you differentiate 1/lna why does it stay as 1/lna?
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    Also remember:
    (a/b)/c = a/(bc)
    a/(b/c) = ac/b

    If you have a fraction at the bottom, you flip it (primary school style), and multiply by the top. If you get confused, try it out with real numbers...
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    (Original post by mockel)
    Also remember:
    (a/b)/c = a/(bc)
    a/(b/c) = ac/b

    If you have a fraction at the bottom, you flip it (primary school style), and multiply by the top. If you get confused, try it out with real numbers...
    safe Mockel! now i wont get so confused.
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    (Original post by Gaz031)
    y=(lna)^-1
    y^-1=(lna)
    (-y^-2)(dy/dx)=1/a
    (dy/dx)=(-1/a)(y^2)
    (dy/dx)=(-1/a)(lna)^-2
    (dy/dx)=-1/[a(lna)^2], which isn't the same as 1/lna.
    Hmm in the edexcel C3 paper A5 it says it stays the same.

    Ok let me say the whole question y = [1/lna] * [lnx]

    When u differentiate it, the answer says it is 1/xlna, and it assumes that when you differetiate 1/lna, it stays the same.
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    (Original post by sb1986)
    Hmm in the edexcel C3 paper A5 it says it stays the same.

    Ok let me say the whole question y = [1/lna] * [lnx]

    When u differentiate it, the answer says it is 1/xlna, and it assumes that when you differetiate 1/lna, it stays the same.
    Actually in my previous post i meant to type x instead of a. I didn't realise you meant a constant rather than a variable, sorry.
    Differentiating 1/lna gives 0, as 1/lna is constant.
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    1/lna is just a constant.
    Differentiating (1/lna)(lnx), is just (1/lna)(1/x) = (1/xlna)

    It's like if you had to do d/dx[3x]. The answer would be 3, since the 3 is similarly, just a constant.
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    yeh, i keep doing these silly mistakes, lol. Thanks
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    Can someone explain the method of finding the range, i know that for a quadratic you can just complete the square, but what about other functions? I know it involves differentiating.
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    Thanks Featherflare, Seth & nTrik. Seems like I forgot to multiply 0.8x by -1. :banghead: :argh:
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    Has anyone managed sin 3x = cos x yet...there was an error in my previous working but the answers I got satisfied the initial equation which is quite a coincidence... :rolleyes:
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    (Original post by sb1986)
    Can someone explain the method of finding the range, i know that for a quadratic you can just complete the square, but what about other functions? I know it involves differentiating.
    Anyone?
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    Nope I think sin3x = cos x is not on the syllabus is it?
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    (Original post by sb1986)
    Anyone?
    Depends on the example - usually drawing a graph is the best first move

    Do you have any specific examples that are troubling you?
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    (Original post by sb1986)
    Anyone?
    Just look at the domain and look at the function, that's how I do it. punch numbers into the calculator to figure out how the curve works and you can generally get it from that.
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    (Original post by sb1986)
    Can someone explain the method of finding the range, i know that for a quadratic you can just complete the square, but what about other functions? I know it involves differentiating.
    For most of them, I draw a graph. ;o
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    (Original post by mik1w)
    Nope I think sin3x = cos x is not on the syllabus is it?
    I'm not sure..someone earlier on in the post said it was on a paper..but I can't seem to solve it properly anyway :confused:
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    More like sin3x ≠ cosx am I right?
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    (Original post by sb1986)
    Can someone explain the method of finding the range, i know that for a quadratic you can just complete the square, but what about other functions? I know it involves differentiating.
    Try drawing a graph as its usually the easiest way to figure it out. The range is the values f(x) or y on the graph can take.
    Alternatively, you can sometimes just figure it out by looking at the domain (this is the values which x can take in the function) and punching numbers into the function.
    If the function is exponential, (eg f(x) = e^a +2 ), then as e^a must be above zero, the range must be f(x) > 0 + 2, f(x) >2

    Hope that helped
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    (Original post by ntrik)
    More like sin3x ≠ cosx am I right?
    Yea iv done the unthinkable.. given up. I dont think it can be done with c1- c4 knowledge which is all i have
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    (Original post by Featherflare)
    Try drawing a graph as its usually the easiest way to figure it out. The range is the values f(x) or y on the graph can take.
    Alternatively, you can sometimes just figure it out by looking at the domain (this is the values which x can take in the function) and punching numbers into the function.
    If the function is exponential, (eg f(x) = e^a +2 ), then as e^a must be above zero, the range must be f(x) > 0 + 2, f(x) >2

    Hope that helped
    Thanks.
 
 
 
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