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    I have a bit of homework and one part of it is to differentiate x^4 from first principles. Obviously the answer is 4x^3 but how do you get this by first principles? I thought I'd learnt this but I'm not sure how to do this with a power above two - does anyone know how it's done?
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    To do it for x^2, you simply take two points on the curve y = x^2, namely (x, x^2) and (x+h, (x+h)^2), then find their gradient, and let h tend to 0. Can you replicate this for x^4?
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    (Original post by generalebriety)
    To do it for x^2, you simply take two points on the curve y = x^2, namely (x, x^2) and (x+h, (x+h)^2), then find their gradient, and let h tend to 0. Can you replicate this for x^4?
    Would it read something like:

    y + \delta\ y = (x + \delta\ x)^4 and then you expand it?

    My mind's gone a bit blank because it's been a long time since I did basic examples of it in C1 :yes:
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    (Original post by Bluebird92)
    Would it read something like:

    y + \delta\ y = (x + \delta\ x)^4 and then you expand it?
    Yes. Do you understand how the differentiation of x^2 works? Just do exactly the same thing - the only different thing is the actual expansion, and you should be able to do that.

    Presumably your method is like this: let y = x^2, and so y + \delta y = (x+\delta x)^2 = x^2 +2x\delta x + (\delta x)^2. So \delta y = (y + \delta y) - y = (x^2 +2x\delta x + (\delta x)^2) - x^2 = 2x\delta x + (\delta x)^2, and so we can easily work out \delta y/\delta x and then take the appropriate limit. Can you do this for x^4?
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    (Original post by generalebriety)
    Yes. Do you understand how the differentiation of x^2 works? Just do exactly the same thing - the only different thing is the actual expansion, and you should be able to do that.

    Presumably your method is like this: let y = x^2, and so y + \delta y = (x+\delta x)^2 = x^2 +2x\delta x + (\delta x)^2. So \delta y = (y + \delta y) - y = (x^2 +2x\delta x + (\delta x)^2) - x^2 = 2x\delta x + (\delta x)^2, and so we can easily work out \delta y/\delta x and then take the appropriate limit. Can you do this for x^4?
    That's what I remember doing for things like x^2 ! The expansion of (x + \delta x)^4 is a bit more troubling to me though.
    For x^2, it is essentially just like completing the square. With x^4 would it be like using Pascal's triangle and so you'd get
    x^4 + 4x^3\delta x + 6x^2\delta x^2 + 4x\delta x^3 + \delta x^4?
    If it was, I could probably go on from there
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    (Original post by Bluebird92)
    That's what I remember doing for things like x^2! The expansion of (x + \deltax)^4 is a bit more troubling to me though.
    For x^2, it is essentially just like completing the square. With x^4 would it be like using Pascal's triangle and so you'd get
    x^4 + 4x^3\delta x + 6x^2\delta x^2 + 4x\delta x^3 + \delta x^4?
    If it was, I could probably go on from there
    Yes, perfect. (If you were really stuck, you could've manually expanded the brackets (x+dx)(x+dx)(x+dx)(x+dx). )
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    (Original post by generalebriety)
    Yes, perfect. (If you were really stuck, you could've manually expanded the brackets (x+dx)(x+dx)(x+dx)(x+dx). )
    Brilliant! I surprised myself there

    Then it's just divide by dy/dx, apply the limits so delta x goes to zero and you're left with 4x^3

    I have to ask though: What if you need to do the same thing with something like \frac{1}{2x+3} ?

    It differentiates to -\frac{2}{(2x+3)^2} using the chain rule which isn't basic differentiation that is seen with terms like x^2 or x^4.

    All I can do is say y + \delta y = \frac{1}{2(x + \delta x) + 3}

    and then \frac{1}{2x + 2\delta x + 3}. But because it's in fraction form, do you need to do anything different with it?

    Thanks for the help so far
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    \displaystyle \frac{1}{2(x+\delta)+3} - \frac{1}{2x+3} = \frac{2x+3 - (2x+3+2\delta)}{(2x+3)(2x+3+2\de  lta)}

    I'll let you finish up from there.
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    (Original post by Bluebird92)
    But because it's in fraction form, do you need to do anything different with it?
    No, not at all. It's always the same principle. DFranklin has answered your question above for this specific function, but the same works for sin(x), ln(x), and so on - it just happens to be a bit more difficult to take the appropriate limit in those cases.
 
 
 
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