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Math Help - Derivitive Question

  1. #1
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    Derivitive Question

    Hi all. I am getting into the chain rule in my Calculus with Matrices business course. I've come across a problem, our system is all online, and i cannt figure it out. They examples they provided were pretty simple and staight forward, but i cannot put htis one together. I'm sure its really easy, but i'd appreciate it if someone would explain

    If f(x)= \frac{1}{X^2 +1}
    then f '(x) is what? The system tell me the answer is \frac{-2X}{(X^2+1)^2}

    I cannot figure out even where to start on this problem nor which derivative rule to use. I'm probably overcomplicating it, but if any one would be willing to help i'd greatly appreciate it!
    Last edited by hpman247; March 22nd 2006 at 11:04 PM.
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  2. #2
    TD!
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    Well, since it's a fraction you could consider the quotient rule, but my advice is to rewrite it as (x^2+1)^{-1} and then just apply the exponent/power rule together with the chain rule. Give it a shot
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  3. #3
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    hey thanks. sorry about the edits, i was playing with the javascript math thing. really neat.

    i'll give it a show
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  4. #4
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    Also, I hate that I forget those basic rules like 1 over a number is the same thing as the denominator of the number to the -1 power. is there a website that i can refesh all that stuff at?

    Thanks again for the help
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  5. #5
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    the "+ 1" is driving me crazy. I dont know waht to do with it. OUr examples didnt tell us what to do if there was any addition or subtraction in the problems w/ the chain rule
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  6. #6
    TD!
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    The derivative is lineair, so the derivative of a sum is equal to the sum of the derivatives.
    The derivative of a constant is 0, so (x+1)' = (x)' + (1)' where ' means derivative, what does that give?
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  7. #7
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    Hi:

    The chain rule is intended to facilitate differentiation of composite functions. In general, if f(g(x)) is the composition of function f with function g, then f '(x) = f '(g) * g '(x). For instance, consider y = e^sin(x). Here y=e^x is composed with u=sin(x). To simplify the thinking, we can write y = e^sin(x) as y = e^u where u=sin(x). So, dy/dx = dy/du * du/dx. Recalling the derivatives of e^x and sin(x) as e^x and cos(x) respectively, it follows that:

    dy/dx = dy/du * du/dx = e^u * cos(x) = e^sin(x) * cos(x)

    In the case at hand, f(x) = (x^2+1)^-1 can be inturpreted as u^-1 where
    u = x^2+1. Hence f '(x) = -1* u^-2 * du/dx = -[x^2+1]^-2 * 2x or,
    f '(x) = -2x / [x^2+1]^2.

    Regards,

    Rich B.
    Last edited by Rich B.; March 23rd 2006 at 02:02 AM.
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