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Posers and Puzzles

Posers and Puzzles

  1. 05 Nov '04 21:59
    i know that the derivative of f(x)=x^2 is f'(x)=2x

    but if we put that in the form of mx+b, where m=the slope, then the coefficent on x^1 is x, so m=x ---->the dreivative.

    clearly this is not the case, by why? it makes a lot of sence

    another example of my point:
    f(x)=x^4-2x^3+3x^2-4x+5
    f'(x)=m=4x^3-6x^2+6x-4
    but in the form mx+b, f(x)=(x^3-2x^2+3x-4)x+5
    so why dosnt m=f'(x)=x^3-2x^2+3x-4 ?
  2. 05 Nov '04 22:42
    Originally posted by fearlessleader
    i know that the derivative of f(x)=x^2 is f'(x)=2x

    but if we put that in the form of mx+b, where m=the slope, then the coefficent on x^1 is x, so m=x ---->the dreivative.

    clearly this is not the case, by why? it makes a lot of sence

    another example of my point:
    f(x)=x^4-2x^3+3x^2-4x+5
    f'(x)=m=4x^3-6x^2+6x-4
    but in the form mx+b, f(x)=(x^3-2x^2+3x-4)x+5
    so why dosnt m=f'(x)=x^3-2x^2+3x-4 ?
    Because that isn't the correct way to differentiate the product of two functions.

    Look up how you differentiate the product of two functions of x and you will see what has gone wrong.
  3. Donation Acolyte
    Now With Added BA
    06 Nov '04 01:59
    Originally posted by fearlessleader
    i know that the derivative of f(x)=x^2 is f'(x)=2x

    but if we put that in the form of mx+b, where m=the slope, then the coefficent on x^1 is x, so m=x ---->the dreivative.

    clearly this is not the case, by why? it makes a lot of sence

    another example of my point:
    f(x)=x^4-2x^3+3x^2-4x+5
    f'(x)=m=4x^3-6x^2+6x-4
    but in the form mx+b, f(x)=(x^3-2x^2+3x-4)x+5
    so why dosnt m=f'(x)=x^3-2x^2+3x-4 ?
    The point is that a derivative measures local rate of change, not some averaged quantity. If we look at x^2 and draw a straight line from the origin to any point, that line will have gradient x for the reason you say. However, the difference between x^2 and (x+h)^2 can be seen as follows: not only do you travel along a steeper slope to get to (x+h)^2 from the origin, you also travel further along it. Effectively this means the h counts twice (as the h^2 term shrinks to nothing for suff. small h), which leads to a factor of 2 in the derivative.
  4. 06 Nov '04 16:28
    i think i understand Acolyte , but if he's in the mood to eloborate, that would be grand.

    i dont have a clue what iamatiger is talking about.
  5. Standard member Palynka
    Upward Spiral
    06 Nov '04 19:10
    Originally posted by fearlessleader
    i think i understand Acolyte , but if he's in the mood to eloborate, that would be grand.

    i dont have a clue what iamatiger is talking about.
    Iamatiger is saying that the form mx+b requires m to be a constant. If m is a function you have to differentiate using the formulas for differentiating products of functions.
  6. 09 Nov '04 21:05
    Originally posted by Palynka
    Iamatiger is saying that the form mx+b requires m to be a constant. If m is a function you have to differentiate using the formulas for differentiating products of functions.
    d/dx [f(x)*g(x)] = f'(x)*g(x) + g'(x)* f(x)

    is the "product rule" for deriving products of two functions...

    however, assume f(x)=k where k is a constant...

    you get

    d/dx[k]*g(x) + g'(x) * k; d/dx[k]= 0 (derivative of a constant is zero)

    so

    0*g(x) + g'(x) * k :. [k*g'(x)] is the result of the product rule when one function - is a constant


    so if you want a concrete example... assume f(x)=k=3 and g(x)=x^2

    d/dx [f(x)*g(x)]

    f'(x)*g(x) + g'(x)*f(x)

    0*x^2 + 2x*3

    6x...

    the mx+b thing only works if m is a constant; the mx+b thing you mentioned is a simplification of the product rule that applies only when m is a constant. if not then that simplification is not valid.
  7. 10 Nov '04 19:05
    i understand.

    i was confusing cause and effect.