anonymous
  • anonymous
totally lost please help, use implicit diffe. to find dy/dx, 2xy+y^2=x+y
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
y(x)1-2y/2x+2y-1
anonymous
  • anonymous
Totally lost sounds bad. Do you know how to differentiate implicitly?.
anonymous
  • anonymous
Lolz totally

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anonymous
  • anonymous
yes thats the answer but again, im lost, @ no data, no sir =(
anonymous
  • anonymous
mon is when my profesor went over this and i was snowed in.
anonymous
  • anonymous
well but maybe you can diferentiante something like this: y = x^2
anonymous
  • anonymous
using dy/dx?
anonymous
  • anonymous
yeah!. and what is the do you get by differentiate y = x^2?
anonymous
  • anonymous
2x/y?
anonymous
  • anonymous
Mmm that was close look: \[\frac{dy}{dx}=2x\]
anonymous
  • anonymous
ok i get that.
anonymous
  • anonymous
It's very important that you know how to get the derivative of explicit functions. By the way, what is the difference between and explict function and an implicit function?
anonymous
  • anonymous
Ok suppose that we have to get the derivative of yx = x^3
anonymous
  • anonymous
You have to use the chain rule. Do you know what rule is that?
anonymous
  • anonymous
yes i do
anonymous
  • anonymous
f'(x)+f(g(x))?
anonymous
  • anonymous
Mmm not exactly.
anonymous
  • anonymous
the chain rule can be written as \[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\]
anonymous
  • anonymous
Let's apply this chain rule to the equation a wrote before.
anonymous
  • anonymous
\[yx=x^3\]
anonymous
  • anonymous
taking the derivative on both sides of the equation.
anonymous
  • anonymous
\[\frac{d}{dx}(yx)=\frac{d}{dx}(x^3)\]
anonymous
  • anonymous
\[\frac{dy}{dx}\frac{dx}{dx}=3x^2\]
anonymous
  • anonymous
is that the answer?
anonymous
  • anonymous
No, it's an example. You need to apply the same rule to your problem.
anonymous
  • anonymous
no i mean to ur example, is that the answer?
anonymous
  • anonymous
yeah.
anonymous
  • anonymous
kk.ima give a go, thanks for the time, i really do appreciate it
anonymous
  • anonymous
You're welcome!

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