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anonymous
 one year ago
find dy/dx of sin(x)=x(1+tan(y))
anonymous
 one year ago
find dy/dx of sin(x)=x(1+tan(y))

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@zepdrix last question of the night :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So whatchu think? Set up that product rule, ya?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so cosx=tany+sec^2(dy/dx) ya?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm \sin(x)=x[1+\tan(y)]\]\[\large\rm \color{royalblue}{[\sin(x)]'}=\color{royalblue}{[x]'}[1+\tan(y)]+x\color{royalblue}{[1+\tan(y)]'}\]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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh! I had it written down but forgot to type it here.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm \color{orangered}{\cos(x)}=\color{orangered}{[1]}[1+\tan(y)]+x\color{royalblue}{[1+\tan(y)]'}\]Orange = Differentiated. Let's take it slow on this last guy

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm \frac{d}{dx}(1+\tan y)=?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 1 would become 0 and the tan y would be sec^2 (y) (dy/dx), is that right?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm \frac{d}{dx}(1+\tan y)=0+\sec^2(y)y'\]Ok great!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm \color{orangered}{\cos(x)}=\color{orangered}{[1]}[1+\tan(y)]+x\color{orangered}{[0+\sec^2(y)y']}\]So we've differentiated all of our stuff successfully. Now we need to solve for y'.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm \cos x=1+\tan y+xy' \sec^2 y\]So how do we do that? :) Any ideas?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you lost me. did you just move the y' to the x in front of sec^2(y)?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Yes :) 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0before you did that I could isolate y' and get (cosxtany1)/(x(sec^2)y) on the other side

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We did a lot like that in class

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is there something wrong with that?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm \cos x=1+\tan y+xy' \sec^2 y\]Subtraction:\[\large\rm \cos x1\tan y=xy' \sec^2 y\]Division:\[\large\rm \frac{\cos x1\tan y}{x \sec^2y}=y'\]Perfect! \c:/

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1I'm not sure what (sec^2)y is. Careful with the way you do your brackets :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0written down it looks identical to yours

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'm just not too good at getting this stuff typed

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Until 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Refer to the attachment using Mathematica v9
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