anonymous
  • anonymous
why does: y+dy=(x+dx)^-2 equal: =x^-2(1+dx/x)^-2
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
put x in evidence
atlas
  • atlas
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
  • atlas
multiplying and dividing Right hand side with x^-2 \[\frac{ x ^{-2}(x+dx)^{-2} }{ x^{-2} }\] \[x^{-2} (\frac{ x+dx }{ x })^{-2}\]

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