Cal 1: Use implicit differentiation to find dy/dx if:

- anonymous

Cal 1: Use implicit differentiation to find dy/dx if:

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- anonymous

|dw:1377722563922:dw|

- 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

Yes I think. I end up getting: y + (x)(dy/dx)?

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- zepdrix

Yup good good good.

- anonymous

Ok great

- zepdrix

So we've gotta deal with that nasty right side huh? :) hmm

- zepdrix

Remember the derivative of lnx?

- anonymous

1/x

- anonymous

So, 1/ tan^-1y?

- 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

So for the right side, |dw:1377723152737:dw| ?

- 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

We would apply it again because of the (1+y^2)?

- anonymous

|dw:1377723314808:dw|

- zepdrix

No we would apply it again because of the y that was inside of our ln[arctan( )].

- 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

Ok yeah I get what you mean. I was going to ask if I needed to add dy/dx!

- anonymous

|dw:1377723422182:dw|

- anonymous

Like that yeah?

- zepdrix

yah looks good.

- zepdrix

So now we need to solve for y'.
\[\Large y+xy'=\frac{1}{(1+y^2)\arctan y}y'\]

- anonymous

So multiply the right denominator to the left side?

- anonymous

for the first step

- 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

What would you have done?

- 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

Hopefully I did that correctly.

- 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

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

They're not the same stuff, maybe I should have called the second group of stuff something else.\[\large (stuff)=y'-(junk)y'\]

- anonymous

I think i understand. Im trying to look at my work and do the same thing.

- 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

Can we fill in the numbers instead of stuff/junk by any chance?

- anonymous

It's hard for me to picture still

- 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

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

Ok I see what you did!

- zepdrix

Then to finish up solving for y' what do we do? :)

- anonymous

Divide?

- zepdrix

Yes good.

- 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

Ohhhhhh I now see what happened to the other dy/dx. Ok, now I am on the same page with you.

- anonymous

You factored it out. It just took me time to see it. lol

- anonymous

I was getting confused, asking myself.. where the other dy/dx went! Now I understand :)

- zepdrix

Ah yes, that step confuses a lot of people! D:

- 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

yay team \c:/

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