clara1223
  • clara1223
find dy/dx of sin(x)=x(1+tan(y))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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clara1223
  • clara1223
@zepdrix last question of the night :)
zepdrix
  • zepdrix
Implicit? :o Oo neato
zepdrix
  • zepdrix
So whatchu think? Set up that product rule, ya?

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clara1223
  • 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?
clara1223
  • 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
\[\large\rm \frac{d}{dx}(1+\tan y)=?\]
clara1223
  • 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?
clara1223
  • 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.
clara1223
  • clara1223
before you did that I could isolate y' and get (cosx-tany-1)/(x(sec^2)y) on the other side
clara1223
  • clara1223
We did a lot like that in class
clara1223
  • 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 :)
clara1223
  • clara1223
written down it looks identical to yours
clara1223
  • 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.
anonymous
  • anonymous
Refer to the attachment using Mathematica v9
1 Attachment

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