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Math Help - Alternative to (x+dx)^-2

  1. #1
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    Alternative to (x+dx)^-2

    The equation is part of the proof of the derivative of a negative exponent in "Calculus Made Easy" by Silvanus P. Thompson.
    The book says (x+dx)^-2 = x^-2(1+dx/x)^-2
    I make it x^-2(1+2.dx/x)^-1

    I know I am wrong but don't know how.

    Note: (dx)^2/x^2 is dropped as it is infintesimal.
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  2. #2
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    Quote Originally Posted by p75213 View Post
    The equation is part of the proof of the derivative of a negative exponent in "Calculus Made Easy" by Silvanus P. Thompson.
    The book says (x+dx)^-2 = x^-2(1+dx/x)^-2
    I make it x^-2(1+2.dx/x)^-1

    I know I am wrong but don't know how.

    Note: (dx)^2/x^2 is dropped as it is infintesimal.
    It looks like the book doesn't drop that term after all.

    \frac{1}{x^2+2xdx+(dx)^2}=\frac{1}{x^2}\ \frac{1}{1+\frac{2dx}{x}+\frac{(dx)^2}{x^2}}

    =\frac{1}{x^2}\ \frac{1}{(1+\frac{dx}{x})^2}
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by p75213 View Post
    The equation is part of the proof of the derivative of a negative exponent in "Calculus Made Easy" by Silvanus P. Thompson.
    The book says (x+dx)^-2 = x^-2(1+dx/x)^-2
    I make it x^-2(1+2.dx/x)^-1

    I know I am wrong but don't know how.

    Note: (dx)^2/x^2 is dropped as it is infintesimal.
    (x+y)^2=x^2+2xy+y^2=x^2(1+2y/x + (y/x)^2)=x^2\left(1+\frac{y}{x}\right)^2

    so:

    \frac{1}{(x+dx)^2}=\frac{1}{x^2\left(1+\frac{dx}{x  }\right)^2}

    There is no dropping of second degree terms (yet)

    CB
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  4. #4
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    Thanks Guys,
    It was quite simple as it turned out, however I could simply not see the solution at the time.
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