2pie
find (dy/dx) by implicit differentiation (x/y)=(y/x)
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anonymous
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|dw:1381786279195:dw|
anonymous
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Quotient rule on both side
2pie
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|dw:1381786349930:dw|
2pie
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is it correct?
what do I do next
2pie
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@iambatman
myininaya
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Good so far.
myininaya
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\[\frac{y-xy'}{y^2}=\frac{xy'-y}{x^2}\]
I would clear the fractions first.
myininaya
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Multiply both sides by x^2y^2.
myininaya
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\[x^2y^2 \frac{y-xy'}{y^2}=x^2y^2\frac{xy'-y}{x^2}\]
You see that we can write this without the fractional style now, right?
2pie
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yes by calcelling the \[y ^{2} x^{2}\] on each side respectively \[x^{2}(y-xy^1)+y^2(xy^1-y)\] then?
myininaya
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Distribute.
Put y' terms on one side and the other non-y' terms on the other side
Then factor y' out.
The divide by whatever is being multiplied by y'.
myininaya
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You mean equal right not plus?
2pie
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yes
2pie
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After I multiply I got\[x^2y-x^3y^\prime =xy^2y^\prime-y^3\]
myininaya
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Right. Continue.
Put y' terms together on one side and then non-y' terms on the other.
2pie
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\[x^2y+y^3=xy^2y^\prime + x^3 y^\prime\]
2pie
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then im not sure what to do
myininaya
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Factor the y' out on that one side.
myininaya
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\[x^2y+y^3=y'(xy^2+x^3)\]
myininaya
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Divide both sides by what y' is being multiplied by.
myininaya
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You can reduce your answer.
2pie
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\[x^2y+y^3/xy^2+x^3\]=\[y^\prime\]
myininaya
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\[y'=\frac{x^2y+y^3}{xy^2+x^3} =\frac{y(x^2+y^2)}{x(y^2+x^2)}\]
Anything cancel?
myininaya
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Now instead of starting the way you guys did at the beginning.
I think @hartnn would like to show you a easier way.
2pie
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ok show me
hartnn
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hey 2pie
since we have x/y = y/x
can we say x^2 = y^2 ?
hartnn
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i just did a cross-multiplication
2pie
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yes
myininaya
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I think the slope can be simplified even more by the way when you guys are done. :)
hartnn
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and then we can differentiate
x^2=y^2
to get 2x =2y dy/dx
can you isolaye dy/dx from this ?
2pie
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yes by dividing 2y on both sides
hartnn
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and what would u get finally ? as dy/dx ?
2pie
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dy/dx = x/y
myininaya
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And recall from your initial equation y/x=x/y
myininaya
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Now another way...
x^2=y^2
or
x^2-y^2=0
or
(x-y)(x+y)=0
or
x-y=0 or x+y=0
or
y=x or y=-x
If y=x, then y'=1
If y=-x, then y'=-1
2pie
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wow it was easier than I thought thank you very much for your help and time @myininaya @hartnn :)
hartnn
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i see you are new here , so
\[ \begin{array}l\color{red}{\text{W}}\color{orange}{\text{e}}\color{#e6e600}{\text{l}}\color{green}{\text{c}}\color{blue}{\text{o}}\color{purple}{\text{m}}\color{purple}{\text{e}}\color{red}{\text{ }}\color{orange}{\text{t}}\color{#e6e600}{\text{o}}\color{green}{\text{ }}\color{blue}{\text{O}}\color{purple}{\text{p}}\color{purple}{\text{e}}\color{red}{\text{n}}\color{orange}{\text{S}}\color{#e6e600}{\text{t}}\color{green}{\text{u}}\color{blue}{\text{d}}\color{purple}{\text{y}}\color{purple}{\text{!}}\color{red}{\text{!}}\color{orange}{\text{ }}\color{#e6e600}{\text{:}}\color{green}{\text{)}}\color{blue}{\text{}}\end{array} \]
2pie
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thank you Im really glad I joined
myininaya
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Who couldn't pass up free math help?
hartnn
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we too are glad that you joined :)
2pie
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thanks ;)