Factoring (x + dx)^{1/2} to get x^{1/2} (1 + dx/x)^{1/2}
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topsquark
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This really shouldn't be such an issue for a Calculus level student.
[imath]\left [ x + dx \right ]^{1/2}[/imath]
[imath]= \left [ x \left ( 1 + \dfrac{dx}{x} \right ) \right ]^{1/2}[/imath]
[imath]= x^{1/2} \left [ 1 + \dfrac{dx}{x} \right ]^{1/2}[/imath]
-Dan
Dr.Peterson
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It isn't clear which part you are asking for help with, or where you are having trouble. Can you tell us more about your difficulties here?
The fact that almost everything written in red is wrong may be relevant. It may also be helpful if you showed more of the context.
This factorization is same pattern as your previous question. Can you show which step needs explanation?
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Try using the *exact* same method as was given to you in your other thread, where the exponent on the *exact* same expression was [imath]-2[/imath] instead of [imath]\frac{1}{2}[/imath].
My guess: book using differentials and binomial expansion instead of limits to discover power rule. Image parts missing, but I see dx^2 and then "terms with higher", so maybe says to ignore terms with exponent>1 in expansion.
y+dy=sqrt[x]*(1+dx/x)^(1/2)
With binomial expansion on right is
y+dy=sqrt[x]+(1/2)*dx/sqrt[x]-(1/8)*dx^2/(x*sqrt[x])+terms with higher exponents
Subtracting original y=x^(1/2) is
dy=(1/2)*dx/sqrt[x]+ignored terms, and that solves for dy/dx to show power rule pattern.
Hi thanks for your reply but next time please pass on your reply. This is a forum to ask questions and I don’t need your additional commentary.
Thanks for the help
The method works on the first problem because once you factor out 2 of the 1/x and then multiply you get 1/x^2 which equal x^-2 . On this problem when you factor out 1/x how does 1/x equal x^1/2 or sqrt x. I see that it was just added but don’t see why it works
BigBeachBanana
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[math](a\cdot b)^c=a^c\cdot b^c[/math]
topsquark
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[imath](ab)^n = a^n b^n[/imath]
My comment and this one aren't meant to be insults: they are observations. If you are having troubles, that's fine, everyone stumbles. But take this as a sign that you need to review.
This should already be in your "bag of tricks" if you are a Calculus student. Knowledge of basic factoring is essential to Calculus.
-Dan
You are mistaken - this is a forum to TEACH. Thus commentaries are inescapable. If you don't want to see them - cover 'em up with a piece of masking tape.
Follow the steps using pencil and paper - don't just stare at it. If you are still confused (after using pencil/paper - take a picture of your most recent work - and we will explain your mistake.
Maybe you misunderstood. In that problem, we factor out x, not 1/x and 1/x.
The 1/x^2 comes because the x factored out is still inside squaring function.
1/[(x)(1+dx/x)]^2=
1/[x]^2 times 1/[1+dx/x]^2
using property of exponents just posted by bigbeachbanana and topsquark
Sorry, I need to leave now, but I can post my explanation in your first thread later.
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Yes the book is explaining where calculations in particular where derivatives come from. I’ve learned and memorized different laws of derivatives but I wanted to know more of the history, theory and postulates behind the math I’ve learned. I feel teacher teach us to memorize and not understand
Yes that works when a*b but that law doesn’t work when it is a+b or when the bases are different. The problem is a+b
So why call the forum free math help if it isn’t to ask questions? What is the point for you to respond saying this forum is hear to teach but then not TEACH!!! Math to me is more than following steps but wondering why it works. Hence probably why you didn’t answer the question. Please pass on my thread and thanks for no help or teaching.
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You asked a question which, in the context of your taking calculus, should not have been a question; you should already have known this. Part of teaching is noticing gaps and pointing them out, so that the student can take corrective action.
You claim that you want to learn the "why" of the process. But you seem not even to have been aware that this exercise is, algebraically, pretty much identical to the other one you posted. Are you saying that you still don't understand "why" that worked the way it did?
Would you prefer, in future, that the helpers just provide you with links to lessons that cover (at least some of) what you don't understand, so we won't be causing you discomfort by trying to have a conversation with you? Perhaps then you won't feel a need to cancel people...?
BigBeachBanana
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This has been shown above but I'll add some more details, maybe it'll help.
[imath]\left [ x + dx \right ]^{1/2}[/imath]
Apply [imath](ab+ac) = a(b+c)[/imath], to factor out the [imath]x[/imath]:
[imath]= \left [ x \left ( 1 + \dfrac{dx}{x} \right ) \right ]^{1/2}[/imath]
Apply exponential rule [imath](a\cdot b)^c=a^c\cdot b^c[/imath]
[imath]= x^{1/2} \left [ 1 + \dfrac{dx}{x} \right ]^{1/2}[/imath]
Ok thanks that’s helps thank you for your patience
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Are you aware that the post you viewed as representing "patience" actually was the opposite? Becoming frustrated with your hostility and refusal to participate in the learning process, somebody finally did all the thinking for you. This is not "math help", but it certainly could explain why you are in calculus without understanding algebra. (For instance, you insisted that [imath](a\cdot b)^c = a^c \cdot b^c[/imath], when the complete worked solution used that very equality.)
Since this thread has become a violation of the terms of use, I will lock this down.
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