Bryanluxor
why does: y+dy=(x+dx)^-2 equal:
=x^-2(1+dx/x)^-2
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RaphaelFilgueiras
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put x in evidence
atlas
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Another way of thinking is divide the expression on R.H.S with x^-2 and then multiply it with x^-2 on the outside
atlas
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multiplying and dividing Right hand side with x^-2
\[\frac{ x ^{-2}(x+dx)^{-2} }{ x^{-2} }\]
\[x^{-2} (\frac{ x+dx }{ x })^{-2}\]
atlas
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simplify it to x^-2 (1+dx/x)^-2
I hope it makes sense now
Bryanluxor
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I stiil dont get it
Could someone please explain it step by step?
myininaya
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Ok, do you agree that x+dx could be written as x(1+dx/x)?
myininaya
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\[x+dx=x+\frac{x}{x} dx=x+x \cdot \frac{dx}{x}=x(1+\frac{dx}{x})\]
Bryanluxor
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how can x+dx be written as x(1+dx/x)?
myininaya
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\[(x+dx)^{-2}=[(x)(1+\frac{dx}{x})]^{-2}=x^{-2}(1+\frac{dx}{x})^{-2} \]
myininaya
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Did you see what I said above what you said?
myininaya
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|dw:1387637203504:dw|
myininaya
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|dw:1387637215111:dw|
myininaya
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you can factor out that x
Bryanluxor
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but how did you get to x+x d/x?
the equation is (x+dx)^-2
myininaya
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You wanted step by step
I'm looking at the inside, x+dx, right now.
Bryanluxor
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ok
myininaya
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If you can understand why x+dx=x(1+dx/x) we can move on to the exponent part
myininaya
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which i already did above
myininaya
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but i can redo
Bryanluxor
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i dont understand why x+dx=x(1+dx/x)
Bryanluxor
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where did you get the 1?
myininaya
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Like do you the distributive property.
It says:
a(b+c)=ab+ac
Bryanluxor
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and how did you get dx/x?
myininaya
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x(1)=x
myininaya
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\[x+dx=x+\frac{x}{x} dx \]
look I multiply dx by x/x which is just one
myininaya
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you should see an x in both terms now
myininaya
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you can factor an x out
Bryanluxor
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how did you get x/x?
myininaya
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\[x(1+\frac{1}{x}dx)\]
\[x(1+\frac{dx}{x}) \]
myininaya
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I just put it there so you would see better
x/x is there if you write it because it is just one
myininaya
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If you factor something out you are dividing the terms you factor from
Bryanluxor
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im sorry but i dont understnad
Bryanluxor
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*understand
myininaya
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For example
\[3x+9=3(\frac{3x}{3}+\frac{9}{3})=3(x+3)\]
Do you understand this example?
Bryanluxor
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Can you please write down every step you do and how you get every variable?
Bryanluxor
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in one answer please
myininaya
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If you don't understand it separately how do you expect to understand it together?
myininaya
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Do you understand the example I just put?
Bryanluxor
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no sorry
Bryanluxor
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im new to calculus
myininaya
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You might want to review factoring then.
Because all I did was factor 3x+9 in that example
Bryanluxor
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i will
myininaya
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Do you know the distributive property?
Bryanluxor
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no sorry
myininaya
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Like a(b+c) means ab+ac
Bryanluxor
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yeah
myininaya
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That is the distributive property
Bryanluxor
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ok
myininaya
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You can look at it either way.
a(b+c)=ab+ac
ab+ac=a(b+c)
Do you see on this side they say hey both of these terms have an a in common so let's factor it out
Bryanluxor
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ok
myininaya
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What they are doing when they factor is pulling something out (multiplying) and then also dividing by what they pulled out.
Like for example ,
ab+ac
we could factor factor out ac but that means we also need to divide each term by ac like this:
\[ac(\frac{ab}{ac}+\frac{ac}{ac})=ac(\frac{b}{c}+1)\]
myininaya
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What do you get when you multiply out the ac(b/c+1) expression?
myininaya
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i must go
good luck
Bryanluxor
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no idea
Bryanluxor
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thankyou
OOOPS
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I don't understand what we are supposed to do, re read your post, y +dy = (x+dx)^-2 equal = x^-2(1+dx/x)^-2
can you use the draw box below to write it again ?
OOOPS
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and dy =y'? dx =x'?
Bryanluxor
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\[y+dy=(x+dx)^{-2} =x ^{-2}(1+\frac{ dx }{ x })^{-2}\]
Bryanluxor
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the aim is to differentiate y=x^2
OOOPS
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2 equal signs?
Bryanluxor
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the second equal sign is the result
Bryanluxor
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once we have powered the left hand side of the equation to -2
Bryanluxor
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what do you mean by "ODE?"
OOOPS
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one more question: from which course , you have this problem?
Bryanluxor
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its from a book called calculus made easy, ill post the page i found it on right now
OOOPS
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thanks for information,
Bryanluxor
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no problem
Bryanluxor
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ill post the page in a sec
Bryanluxor
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Bryanluxor
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he adds dy or dx to expand it a bit
OOOPS
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I can understand (x +dx)^(-2) = x^(-2)(1+dx/x)^-2
but from y +dy , I cannot see how they get the right hand side.
since y = x^-2 --> dy = -2x^-3
y + dy = x^-2 - 2 x^-3 = x^-2(1-2dx/x) which is not their right hand side :((
zpupster
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this might clear it up
i like this book
his proofs are easier to understand than
the way was taught.