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

- anonymous

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

- schrodinger

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- anonymous

put x in evidence

- 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

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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## More answers

- atlas

simplify it to x^-2 (1+dx/x)^-2
I hope it makes sense now

- anonymous

I stiil dont get it
Could someone please explain it step by step?

- myininaya

Ok, do you agree that x+dx could be written as x(1+dx/x)?

- myininaya

\[x+dx=x+\frac{x}{x} dx=x+x \cdot \frac{dx}{x}=x(1+\frac{dx}{x})\]

- anonymous

how can x+dx be written as x(1+dx/x)?

- myininaya

\[(x+dx)^{-2}=[(x)(1+\frac{dx}{x})]^{-2}=x^{-2}(1+\frac{dx}{x})^{-2} \]

- myininaya

Did you see what I said above what you said?

- myininaya

|dw:1387637203504:dw|

- myininaya

|dw:1387637215111:dw|

- myininaya

you can factor out that x

- anonymous

but how did you get to x+x d/x?
the equation is (x+dx)^-2

- myininaya

You wanted step by step
I'm looking at the inside, x+dx, right now.

- anonymous

ok

- myininaya

If you can understand why x+dx=x(1+dx/x) we can move on to the exponent part

- myininaya

which i already did above

- myininaya

but i can redo

- anonymous

i dont understand why x+dx=x(1+dx/x)

- anonymous

where did you get the 1?

- myininaya

Like do you the distributive property.
It says:
a(b+c)=ab+ac

- anonymous

and how did you get dx/x?

- myininaya

x(1)=x

- myininaya

\[x+dx=x+\frac{x}{x} dx \]
look I multiply dx by x/x which is just one

- myininaya

you should see an x in both terms now

- myininaya

you can factor an x out

- anonymous

how did you get x/x?

- myininaya

\[x(1+\frac{1}{x}dx)\]
\[x(1+\frac{dx}{x}) \]

- 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

If you factor something out you are dividing the terms you factor from

- anonymous

im sorry but i dont understnad

- anonymous

*understand

- myininaya

For example
\[3x+9=3(\frac{3x}{3}+\frac{9}{3})=3(x+3)\]
Do you understand this example?

- anonymous

Can you please write down every step you do and how you get every variable?

- anonymous

in one answer please

- myininaya

If you don't understand it separately how do you expect to understand it together?

- myininaya

Do you understand the example I just put?

- anonymous

no sorry

- anonymous

im new to calculus

- myininaya

You might want to review factoring then.
Because all I did was factor 3x+9 in that example

- anonymous

i will

- myininaya

Do you know the distributive property?

- anonymous

no sorry

- myininaya

Like a(b+c) means ab+ac

- anonymous

yeah

- myininaya

That is the distributive property

- anonymous

ok

- 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

ok

- 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

What do you get when you multiply out the ac(b/c+1) expression?

- myininaya

i must go
good luck

- anonymous

no idea

- anonymous

thankyou

- 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

and dy =y'? dx =x'?

- anonymous

\[y+dy=(x+dx)^{-2} =x ^{-2}(1+\frac{ dx }{ x })^{-2}\]

- anonymous

the aim is to differentiate y=x^2

- anonymous

2 equal signs?

- anonymous

the second equal sign is the result

- anonymous

once we have powered the left hand side of the equation to -2

- anonymous

what do you mean by "ODE?"

- anonymous

one more question: from which course , you have this problem?

- anonymous

its from a book called calculus made easy, ill post the page i found it on right now

- anonymous

thanks for information,

- anonymous

no problem

- anonymous

ill post the page in a sec

- anonymous

##### 1 Attachment

- anonymous

he adds dy or dx to expand it a bit

- 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

this might clear it up
i like this book
his proofs are easier to understand than
the way was taught.

##### 1 Attachment

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