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Math Help - derivative of y with respect to (1-x)

  1. #1
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    derivative of y with respect to (1-x)

    I need help with differentiation.
    I don't quite understand what is derivative of y with respect to some function. Yes x is also a function but let me show the example.
    if y = x^2 + 2, what is derivative of y with respect to (1-x) ?
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  2. #2
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    Question

    I've never heard of such a thing. Are you perhaps supposed to do some sort of substitution, like "u = 1 - x"? So then x = 1 - u, and x^2 + 2 = u^2 - 2u + 3? And then you differentiate with respect to u, and then maybe back-substitute?

    I'm just guessing....
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  3. #3
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    Quote Originally Posted by tabularasa View Post
    I need help with differentiation.
    I don't quite understand what is derivative of y with respect to some function. Yes x is also a function but let me show the example.
    if y = x^2 + 2, what is derivative of y with respect to (1-x) ?
    You can differentiate any (differentiable) function with respect to any other (differentiable) function. Let f(x) and g(x) be differentiable functions of x. By the chain rule, \frac{df}{dg}= \frac{df}{dx}\frac{dx}{dg} = \frac{df}{dx}\frac{1}{\frac{dg}{dx}}= \frac{\frac{df}{dx}}{\frac{dg}{dx}}.

    Here f(x)= x^2+ 2 and g(x)= 1- x. \frac{df}{dx}= 2x and \frac{dg}{dx}= -1 so \frac{d(x^2+2)}{d(1-x)}= \frac{df}{dg}= \frac{2x}{-1}= -2x.

    The derivative of f with respect to g essentially tells how fast f is increasing or decreasing compared to how fast g is.
    Last edited by HallsofIvy; November 12th 2009 at 04:24 AM.
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  4. #4
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    Exactly! Thanks.

    I substitute and back subsitute.

    \frac{d}{d(x-1)} y
    y = x^2+1

    and then substituting:
    u = x-1, x=1+u

    and with chain rule I get:
    \frac{d}{du} (u+1)^2+1 = D((u+1)^2) \times D(u+1) = 2(u+1)\times1

    back substitute x:
    2(1-x)+1) = 2-4x

    Thanks but hmm...you get -2x ?
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  5. #5
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    In your original post you said you were differentiating with respect to 1-x. In your last post you used x- 1. Did you change problems?
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    In your original post you said you were differentiating with respect to 1-x. In your last post you used x- 1. Did you change problems?
    Oh, damn. Original 1-x is right question. I update my equations. Thanks!
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  7. #7
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    EQUATION UPDATED. My mistake earlier.

    I substitute and back subsitute.

    \frac{d}{d(1-x)} y
    y = x^2+1

    and then substituting:
    u = 1-x, x=1-u

    and with chain rule I get:
    \frac{d}{du} (1-u)^2+1 = D((1-u)^2) \times D(1-u) = 2(1-u)\times1

    back substitute x:
    2(1-1-x) = -2x


    HallsofIvy, your example is much more faster, I have to study that.

    Thank you guys!!
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