Algebra question

  • Thread starter JT73
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  • #1
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I watched some of the first video on the calculus lectures someone posted, but had a question about some of the algebra when he factors something out of the equation. I've noticed stuff like this is where I get messed up. If someone could explain how this equation:
[1/(x + deltax) - 1/(x)] / deltax

turns into

(1/deltax)[(x - (x + deltax)) /(x + delta x)(x)]

The "X"s, not the "deltaXs", were written with a little subscript zero to the bottom of them, but I don't know how to do that on my keyboard, my apologies. Also, it may hard to read the way it is written, but hopefully you guys can get the jist of it. Thanks
 
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Answers and Replies

  • #2
1,065
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First, instead of having a numerator and denominator fraction where the numerator itself consists of fractions with their own numerator and denominator...you need to take the denominator and pull it out (1/delta) as multiplying the whole numerator...

...now, you should have the multiplication of two factors: (1/delat) and the rest.

then, you take the second factor and join them using a common denominator.
 
  • #3
Deveno
Science Advisor
906
6
[tex]\frac{\frac{1}{x_0 + \Delta x}-\frac{1}{x_0}}{\Delta x} = \left(\frac{1}{\Delta x}\right)\left(\frac{1}{x_0 + \Delta x}-\frac{1}{x_0}\right)[/tex]

[tex]= \left(\frac{1}{\Delta x}\right)\left( \frac{x_0}{(x_0+\Delta x)(x_0)} - \frac{x_0+\Delta x}{(x_0+\Delta x)(x_0)}\right) [/tex]

[tex]= \left(\frac{1}{\Delta x}\right)\left( \frac{x_0 - (x_0 + \Delta x)}{(x_0 + \Delta x)(x_0)}\right) = \left(\frac{1}{\Delta x}\right)\left( \frac{-\Delta x}{(x_0 + \Delta x)(x_0)}\right)[/tex]

[tex]= \frac{-1}{(x_0 + \Delta x)(x_0)}[/tex]

commentary:

in the first step, we note that division by u is the same thing as multiplication by 1/u (here our "u" is "delta x").

next, we multiply each fraction in the difference, by the denominator of the other fraction in the difference, to get a common denominator:

(a/b - c/d) = ((a/b)(1) - (c/d)(1)) = ((a/b)(d/d) - (c/d)(b/b)) = ((ad)/(bd) - (bc)/(bd)) <--common denominator of bd

once we have a common denominator, we can subtract numerators:

(ad)/(bd) - (bc)/(bd) = (ad - bc)/(bd) (this is just the standard rule for subtracting fractions).

once that is done, we cancel the "delta x's".
 
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  • #4
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Oh man, you have my sincerest thanks. That is exactly what I needed to see. The ideas and concepts of most math that I have taken and studied on my own (calculus) do not seem to hard to me. It is just things I do not have down that well, here and there, that cause roadblocks. Like my initial question.
 

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