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find (dy/dx) by implicit differentiation (x/y)=(y/x)

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Quotient rule on both side

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Other answers:

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

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