Factoring: How does (x+dx)^-2 = x^-2(1+dx/x)^-2 ?
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If I understand correctly, you have the following equation:
[imath]\dfrac{1}{(x+dx)^2} = \left(\dfrac{1}{x^2}\right)\left(\dfrac{1}{(1 + \frac{dx}{x})^2}\right)[/imath]
...and you're wanting to figure out how the left-hand side (LHS) equals the right-hand side (RHS).
Hint: Look at what is different on the RHS, and see if you can figure out how that relates to the LHS. What did they factor out front? What would this leave you with?
The Highlander
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Hi @jmoney30,
Have you figured this out yet?
You can show how the LHS equals the RHS by using some (fairly) simple algebraic manipulation.
If you're not sure how to go about that, I will give you some starters that should enable you to do so...
Your initial equation (as helpfully provided by @stapel) is:-
[math]\dfrac{1}{(x+dx)^2} = \left(\dfrac{1}{x^2}\right)\left(\dfrac{1}{(1 + \frac{dx}{x})^2}\right)[/math]
Now, multiplying the two (bracketted) fractions on the RHS, you get:-
[math]\left(\dfrac{1}{x^2}\right)\left(\dfrac{1}{(1 + \frac{dx}{x})^2}\right) = \dfrac{1\times 1}{x^2\times (1 + \frac{dx}{x})^2}=\dfrac{1}{x^2(1 + \frac{dx}{x})^2}[/math]
[math]=\dfrac{1}{x^2(1 + d)^2}\text{ (since }\scriptsize\frac{dx}{x}\normalsize\text{ just equals }d)[/math]
So now you have:- [math]\dfrac{1}{(x+dx)^2}=\dfrac{1}{x^2(1 + d)^2}[/math]
and, using the very useful suggestion provided by @Steven G, you can then say...
[math] \dfrac{1}{(x+dx)^2}=\dfrac{1}{x^2(1 + d)^2}[/math]
[math]\implies (x+dx)^2=x^2(1 + d)^2[/math]
[math]\implies (x+dx)^2=x^2(1 + d)^2[/math]
And now all that's left for you to do is to expand the brackets, which I trust you know how to do, for example:-
[math](1+d)^2=(1+d)\times (1+d)=(1^2+d+d+d^2)=(1+2d+d^2)[/math]
followed by a little bit of multiplication (throughout the RHS) to show that
[math]\dfrac{1}{(x+dx)^2}\text{ does, indeed, equal }\left(\dfrac{1}{x^2}\right)\left(\dfrac{1}{(1 + \frac{dx}{x})^2}\right)[/math]
Please now come back and show us how you would complete this exercise.
NB: You don't have to produce all the fancy formatting of the equations (as shown above), just write your work out (in pencil?) on a sheet of paper or in your jotter then take a picture of your completed work and post that as your reply.
Hope that helps. ?
(And we look forward to hearing from you again soon. ?)
Last edited:
Thanks I figured it out yesterday about a 30 minutes after I posted . It’s converting the power than factoring out from the squared frunction then converted the power. My terminology might be wrong but I took it step by step. Thanks
Dr.Peterson
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I'm going to guess that the question is not how to prove that this is true, but how to get from the LHS to the RHS (and, perhaps, why).
We could more easily see (and show) why you'd want to do this if we saw the context in which it is done. An image of the source would help.
As to how, supposing that part of the goal is to make it look like [imath](1+a)^n[/imath], I would start by factoring [imath]x[/imath] out of [imath](x+dx)[/imath], by dividing each term by [imath]x[/imath], and writing it outside: [imath](x+dx)=x(1+\frac{dx}{x})[/imath]. Then just raise that to the -2 power.
This sort of thing often confuses students because they are accustomed to factoring as taking out a visible common factor, and not involving fractions. This is a different use of factoring, which is typically to make an expression take a certain form.
Steven G
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I am confused about two things that you did?
You want to show that \(\displaystyle (x+dx)^2=x^2(1 + d)^2\)
I just don't get why you chose to multiply out (1+d)^2? Why not use the fact that a^n*b^n = (ab)^n
The 2nd point that totally confuses me is how does dx/x = d? After all, \(\displaystyle dx \neq d*x\). Others say that you're work is correct but I don't see how. I'm not saying that you're wrong but rather am asking if I'm missing something?
The Highlander
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I'm very sorry that I confused you (again, lol), Steven; my sincerest apologies. ?
I always (or, at least, usually) try to aim my help at the least "expert" visitors who might view any of my posts or are likely to have any interest in the topic being discussed as that (IMNHO ?) benefits the widest audience. ?
I also chose to expand (1 + d)2 (rather than (x + dx)2) because I had already done 90% of the work for the OP and wanted to leave at least something for him/her to do!
In retrospect (it occurred to me only after it was to late to edit the post) I felt I should have actually expanded something like (1 + x)2 or even (1 + a)2 rather than using one of the expressions that was actually included in the equation since I was only providing an example of how to expand a (squared) bracket term.
(Let's not open up that old can of worms again; you may recall that we already wasted enough time and forum space fighting over the best way to approach that technique in the distant past! lol)
* When s/he wrote: "It’s a factoring problem not solving for x or derivative"
Hope that clears things up for you. ?
I multiplied out (1 + d)2 simply as an example of how to expand brackets (you will note that I even put in steps that we would normally just omit too!) because I was aiming my answer at the "lowest common denominator"; ie so that a beginner in algebra would understand what I was talking about (for the benefit of, not only the OP (if indeed s/he needed such help) but also for anyone else viewing the thread at a later date who was just coming to terms with basic algebraic manipulation). ?
I always (or, at least, usually) try to aim my help at the least "expert" visitors who might view any of my posts or are likely to have any interest in the topic being discussed as that (IMNHO ?) benefits the widest audience. ?
I also chose to expand (1 + d)2 (rather than (x + dx)2) because I had already done 90% of the work for the OP and wanted to leave at least something for him/her to do!
In retrospect (it occurred to me only after it was to late to edit the post) I felt I should have actually expanded something like (1 + x)2 or even (1 + a)2 rather than using one of the expressions that was actually included in the equation since I was only providing an example of how to expand a (squared) bracket term.
(Let's not open up that old can of worms again; you may recall that we already wasted enough time and forum space fighting over the best way to approach that technique in the distant past! lol)
I believe the OP made it clear* in the first post that the term dx was not supposed to be an example of Leibniz's notation but was, indeed, just supposed to represent d × x, otherwise the problem wouldn't make any sense at all, would it? (If the dx was Leibniz's notation.) ?
* When s/he wrote: "It’s a factoring problem not solving for x or derivative"
Hope that clears things up for you. ?
The Highlander
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Also (meant to say) , although it's perfectly true that an × bn = (a × b)n, pre algebra students (or the general public) may well not be at all familiar with that identity and could well then be left perplexed by x2(1 + d)2 'suddenly becoming' (x + dx)2 with no explanation as to how that came about. ?
That is another reason why I was leaving it for the OP to do (some) work by finishing the problem explicitly by expanding the brackets & multiplying throughout the RHS by the x2 term thus...
\(\displaystyle (x+dx)^2=x^2(1+d)^2\)
\(\displaystyle \implies (x^2+2dx^2+d^2x^2)=x^2(1+2d+d^2)\)
\(\displaystyle \implies (x^2+2dx^2+d^2x^2)=(x^2+2dx^2+d^2x^2)\) QED. ?
Which anyone with even a smattering of algebraic understanding can (surely) follow. Huh?
Steven G
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The fact that the OP even mentioned Calculus was suspicious to me. Many times a student in Calculus will post their question in say algebra because they feel that their question is an algebra question vs a calculus question.
Also, the dx/x part was suspiciously easy for me to think non calculus.
Also, the dx/x part was suspiciously easy for me to think non calculus.
Dr.Peterson
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I hope you recognize from the other thread that this is about calculus:
Factoring (x + dx)^{1/2} to get x^{1/2} (1 + dx/x)^{1/2}
Hi I found another one from a book I’m reading. What are the steps to factor. thanks
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Steven G
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Factoring: How does (x+dx)^-2 = x^-2(1+dx/x)^-2 ?
This is equivalent to (x+dx)^2 = x^2(1+dx/x)^2.Now the lhs becomes x^2 + 2xdx + (dx)^2 = 1(x^2) +x^2(2dx/x) + x^2(dx/x)^2 = x^2(1+2dx/x+ (dx/x)^2) = x^2( 1+ dx/x)^2.
I didn't think that this could be done with calculus (like not using dx/x = d) but I did do it.
What I have above was basically my scratch work and since it worked out I decided to leave it here. I hope everyone is ok with my posting a complete solution.
Dr.Peterson
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Sure.
Just as the other thread is about differentiating [imath]y=x^{1/2}[/imath],
this one is about differentiating [imath]y=x^{-2}[/imath]. We want to find (approximate) dy for a given dx, by finding that [imath]y + dy = f(x + dx) = (x + dx)^{-2}[/imath], and then (though we haven't been shown exactly how they choose to do it) using the generalized binomial theorem and, probably, neglecting terms with higher powers of dx.
The Highlander
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I can't (as you can see, from my "explanations" above, I interpreted it as a purely (Pre-) Algebra problem) but it's good to see that the venerable @Dr.Peterson can (and has done)! ?
However, could I ask (everyone?) to write x2 rather than x^2. It only takes a moment longer (see attached gif image, down below) and makes it so much easier to read (decipher?) the likes of, just for example, ...
x2 + 2xdx + (dx)2 = 1(x2) + x2(2dx/x) + x2(dx/x)2 = x2(1 + 2dx/x + (dx/x)2) = x2(1 + dx/x)2
as opposed to...
x^2 + 2xdx + (dx)^2 = 1(x^2) +x^2(2dx/x) + x^2(dx/x)^2 = x^2(1+2dx/x+ (dx/x)^2) = x^2( 1+ dx/x)^2
Wouldn't you (all?) agree?
It is not, of course, necessary to italicize the variable (I just like to do it because it makes it look more like it would if LaTex had been used). ?
Hope that's of interest (to at least some). ?
I agree with Dr. Peterson. If you would like to see example, I post it in other thread.
Factoring (x + dx)^{1/2} to get x^{1/2} (1 + dx/x)^{1/2}
Hi I found another one from a book I’m reading. What are the steps to factor. thanks
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Dr.Peterson
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In that case, let's show the proper work for the question, which was
[math](x+dx)^{-2} = \left(x\left(1+\dfrac{dx}{x}\right)\right)^{-2} =x^{-2}\left(1+\dfrac{dx}{x}\right)^{-2}[/math]
That's all it takes.
The next part, which wasn't asked about and is presumably in the book, was nicely done by @Bruce for the 1/2 power, which I'll format here, since I couldn't do it there:
Redoing that for our problem:
Although the pieces of the work have been shown here and there, the direct way is this:
[math](x+dx)^{-2} = \left(x\left(1+\dfrac{dx}{x}\right)\right)^{-2} =x^{-2}\left(1+\dfrac{dx}{x}\right)^{-2}[/math]
That's all it takes.
The next part, which wasn't asked about and is presumably in the book, was nicely done by @Bruce for the 1/2 power, which I'll format here, since I couldn't do it there:
Redoing that for our problem:
[imath]y+dy=x^{-2}\left(1+\dfrac{dx}{x}\right)^{-2}[/imath]
[imath]y+dy=x^{-2}\left(1-2\dfrac{dx}{x}+3\dfrac{dx^2}{x^2}+\text{ignored terms}\right)=x^{-2}-2\dfrac{dx}{x^3}+3\dfrac{dx^2}{x^4}+\text{ignored terms}[/imath]
Subtracting original [imath]y=x^{-2}[/imath], [imath]dy=-2\dfrac{dx}{x^3}+3\dfrac{dx^2}{x^4}+\text{ignored terms}[/imath]
Ignoring the higher-degree terms and dividing by dx, dy/dx is approximately [imath]\dfrac{-2}{x^3}[/imath], which of course is correct.