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why does: y+dy=(x+dx)^-2 equal: =x^-2(1+dx/x)^-2

Mathematics
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put x in evidence
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
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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Other answers:

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