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Mathematics 9 Online
OpenStudy (clara1223):

find dy/dx of sin(x)=x(1+tan(y))

OpenStudy (clara1223):

@zepdrix last question of the night :)

zepdrix (zepdrix):

Implicit? :o Oo neato

zepdrix (zepdrix):

So whatchu think? Set up that product rule, ya?

OpenStudy (clara1223):

so cosx=tany+sec^2(dy/dx) ya?

zepdrix (zepdrix):

\[\large\rm \sin(x)=x[1+\tan(y)]\]\[\large\rm \color{royalblue}{[\sin(x)]'}=\color{royalblue}{[x]'}[1+\tan(y)]+x\color{royalblue}{[1+\tan(y)]'}\]Hmm I feel like you're missing an x on the right side. See my product rule set up? See how I still have an x in one of the terms on the right?

OpenStudy (clara1223):

Oh! I had it written down but forgot to type it here.

zepdrix (zepdrix):

\[\large\rm \color{orangered}{\cos(x)}=\color{orangered}{[1]}[1+\tan(y)]+x\color{royalblue}{[1+\tan(y)]'}\]Orange = Differentiated. Let's take it slow on this last guy

zepdrix (zepdrix):

\[\large\rm \frac{d}{dx}(1+\tan y)=?\]

OpenStudy (clara1223):

the 1 would become 0 and the tan y would be sec^2 (y) (dy/dx), is that right?

zepdrix (zepdrix):

\[\large\rm \frac{d}{dx}(1+\tan y)=0+\sec^2(y)y'\]Ok great!

zepdrix (zepdrix):

\[\large\rm \color{orangered}{\cos(x)}=\color{orangered}{[1]}[1+\tan(y)]+x\color{orangered}{[0+\sec^2(y)y']}\]So we've differentiated all of our stuff successfully. Now we need to solve for y'.

zepdrix (zepdrix):

\[\large\rm \cos x=1+\tan y+xy' \sec^2 y\]So how do we do that? :) Any ideas?

OpenStudy (clara1223):

you lost me. did you just move the y' to the x in front of sec^2(y)?

zepdrix (zepdrix):

Yes :) The y' is outside of the secant function. So I just moved in front. Too many brackets, was trying to clean things up a tad.

OpenStudy (clara1223):

before you did that I could isolate y' and get (cosx-tany-1)/(x(sec^2)y) on the other side

OpenStudy (clara1223):

We did a lot like that in class

OpenStudy (clara1223):

Is there something wrong with that?

zepdrix (zepdrix):

\[\large\rm \cos x=1+\tan y+xy' \sec^2 y\]Subtraction:\[\large\rm \cos x-1-\tan y=xy' \sec^2 y\]Division:\[\large\rm \frac{\cos x-1-\tan y}{x \sec^2y}=y'\]Perfect! \c:/

zepdrix (zepdrix):

I'm not sure what (sec^2)y is. Careful with the way you do your brackets :)

OpenStudy (clara1223):

written down it looks identical to yours

OpenStudy (clara1223):

i'm just not too good at getting this stuff typed

zepdrix (zepdrix):

oh i see :D

zepdrix (zepdrix):

Until you become a \(\large \color{orange}{master}\) at these cool little rules, I would strongly recommend you `set up your product rules before doing them`. I should take that back actually.. If you're the type of student who is prone to making lots of little errors, then maybe the less you write down the better. I've known quite a few students who are like that. But personally I really like the set up.

OpenStudy (anonymous):

Refer to the attachment using Mathematica v9

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