Algebra: Solving Equation with DeltaXs

[JT73]

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

 

[gsal]

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.

 

[Deveno]

$$\frac{\frac{1}{x_0 + \Delta x}-\frac{1}{x_0}}{\Delta x} = \left(\frac{1}{\Delta x}\right)\left(\frac{1}{x_0 + \Delta x}-\frac{1}{x_0}\right)$$

$$= \left(\frac{1}{\Delta x}\right)\left( \frac{x_0}{(x_0+\Delta x)(x_0)} - \frac{x_0+\Delta x}{(x_0+\Delta x)(x_0)}\right)$$

$$= \left(\frac{1}{\Delta x}\right)\left( \frac{x_0 - (x_0 + \Delta x)}{(x_0 + \Delta x)(x_0)}\right) = \left(\frac{1}{\Delta x}\right)\left( \frac{-\Delta x}{(x_0 + \Delta x)(x_0)}\right)$$

$$= \frac{-1}{(x_0 + \Delta x)(x_0)}$$

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".

 

[JT73]

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.

 

[CellCoree]

I understand that algebra can be confusing and challenging at times. It is important to have a solid understanding of algebra in order to solve equations and understand mathematical concepts, especially in fields such as calculus.

In this equation, we are trying to simplify the expression by factoring. The first thing to note is that the denominator, deltax, is common to both fractions. Therefore, we can factor it out and rewrite the equation as:

1/deltax * (1/(x+deltax) - 1/x)

Next, we can simplify the fractions by finding a common denominator. In this case, the common denominator is (x+deltax)(x). We can rewrite the fractions with this common denominator as:

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

Now, we can combine the fractions by subtracting the numerators and keeping the common denominator:

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

Finally, we can simplify the numerator by distributing the negative sign and simplifying the expression to get the final answer:

1/deltax * [(x - x - deltax)/((x+deltax)(x))] = 1/deltax * (-deltax/((x+deltax)(x))) = (-1/(x+deltax)(x))

I hope this explanation helps to clarify how the equation was simplified. Remember to always look for common factors and common denominators when simplifying algebraic expressions. Practice makes perfect, so keep practicing and you will become more comfortable with algebraic manipulations.