anonymous
  • anonymous
Cal 1: Use implicit differentiation to find dy/dx if:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1377722563922:dw|
zepdrix
  • zepdrix
Do you understand how to approach the left side? Looks like we'll have to apply the product rule:\[\Large (xy)'\qquad=\qquad \color{royalblue}{(x)'}y+x\color{royalblue}{(y)'}\]
anonymous
  • anonymous
Yes I think. I end up getting: y + (x)(dy/dx)?

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zepdrix
  • zepdrix
Yup good good good.
anonymous
  • anonymous
Ok great
zepdrix
  • zepdrix
So we've gotta deal with that nasty right side huh? :) hmm
zepdrix
  • zepdrix
Remember the derivative of lnx?
anonymous
  • anonymous
1/x
anonymous
  • anonymous
So, 1/ tan^-1y?
zepdrix
  • zepdrix
Yes good good. Just don't forget the chain rule!\[\Large \left[\ln(\arctan y)\right]' \qquad=\qquad \frac{1}{\arctan y}\color{royalblue}{\left(\arctan y\right)'}\]
anonymous
  • anonymous
So for the right side, |dw:1377723152737:dw| ?
zepdrix
  • zepdrix
Good but we need to apply the chain rule AGAIN! :) lol\[\Large \frac{1}{\arctan y}\cdot\frac{1}{1+y^2}\color{royalblue}{(y)'}\]
anonymous
  • anonymous
We would apply it again because of the (1+y^2)?
anonymous
  • anonymous
|dw:1377723314808:dw|
zepdrix
  • zepdrix
No we would apply it again because of the y that was inside of our ln[arctan( )].
zepdrix
  • zepdrix
If that confuses you, you don't need to think of that step as the chain rule (even though that's what we're doing). You can just remind yourself, "Oh the variable is y, I need to attach a dy/dx to the end of it"
anonymous
  • anonymous
Ok yeah I get what you mean. I was going to ask if I needed to add dy/dx!
anonymous
  • anonymous
|dw:1377723422182:dw|
anonymous
  • anonymous
Like that yeah?
zepdrix
  • zepdrix
yah looks good.
zepdrix
  • zepdrix
So now we need to solve for y'. \[\Large y+xy'=\frac{1}{(1+y^2)\arctan y}y'\]
anonymous
  • anonymous
So multiply the right denominator to the left side?
anonymous
  • anonymous
for the first step
zepdrix
  • zepdrix
Ummm ya sure, that might clean up the problem a little bit. Make it look a lot nicer. That's not really the step we want to execute in order to solve for y'. But it's a good intermediate step to make this look a bit friendlier.
anonymous
  • anonymous
What would you have done?
zepdrix
  • zepdrix
No it's a good step, let's do that first :) It gets rid of the ugly fraction.\[\Large (y+y^3)\arctan y+x(1+y^2)(\arctan y)y'=y'\]
zepdrix
  • zepdrix
Hopefully I did that correctly.
zepdrix
  • zepdrix
Think of the equation something like this,\[\large (stuff)+(stuff)y'=y'\] To solve for y', we need add/subtract to get both terms on the same side.
zepdrix
  • zepdrix
So let's subtract the one with stuff on it, from each side,\[\large (stuff)=y'-(stuff)y'\] Understand what I'm doing? Or is the (stuff) making it confusing? lol
zepdrix
  • zepdrix
They're not the same stuff, maybe I should have called the second group of stuff something else.\[\large (stuff)=y'-(junk)y'\]
anonymous
  • anonymous
I think i understand. Im trying to look at my work and do the same thing.
zepdrix
  • zepdrix
The reason we want them all on the same side is so we can `factor` from here. We'll factor a y' from each term on the right.\[\Large (stuff)=y'\left[1-(junk)\right]\]
anonymous
  • anonymous
Can we fill in the numbers instead of stuff/junk by any chance?
anonymous
  • anonymous
It's hard for me to picture still
zepdrix
  • zepdrix
\[\Large (y+y^3)\arctan y+x(1+y^2)(\arctan y)y'\quad=\quad y'\]Subtracting x(1+y^2)(arctany)y' from each side gives us,\[\Large (y+y^3)\arctan y\quad=\quad y'-x(1+y^2)(\arctan y)y'\]
zepdrix
  • zepdrix
Factoring a y' out of each term gives us,\[\Large (y+y^3)\arctan y\quad=\quad y'\left[1-x(1+y^2)(\arctan y)\right]\]
anonymous
  • anonymous
Ok I see what you did!
zepdrix
  • zepdrix
Then to finish up solving for y' what do we do? :)
anonymous
  • anonymous
Divide?
zepdrix
  • zepdrix
Yes good.
zepdrix
  • zepdrix
Ending up with something like this I guess? :O Kinda scary huh? lol\[\Large y'=\frac{(y+y^3)\arctan y}{1-x(1+y^2)\arctan y}\]
anonymous
  • anonymous
Ohhhhhh I now see what happened to the other dy/dx. Ok, now I am on the same page with you.
anonymous
  • anonymous
You factored it out. It just took me time to see it. lol
anonymous
  • anonymous
I was getting confused, asking myself.. where the other dy/dx went! Now I understand :)
zepdrix
  • zepdrix
Ah yes, that step confuses a lot of people! D:
anonymous
  • anonymous
Thank you so much for spending your time helping me. I feel like I know how to solve it, I just needed guidance.
zepdrix
  • zepdrix
yay team \c:/

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