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clara1223

  • one year ago

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

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  1. clara1223
    • one year ago
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    @zepdrix last question of the night :)

  2. zepdrix
    • one year ago
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    Implicit? :o Oo neato

  3. zepdrix
    • one year ago
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    So whatchu think? Set up that product rule, ya?

  4. clara1223
    • one year ago
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    so cosx=tany+sec^2(dy/dx) ya?

  5. zepdrix
    • one year ago
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    \[\large\rm \sin(x)=x[1+\tan(y)]\]\[\large\rm \color{royalblue}{[\sin(x)]'}=\color{royalblue}{[x]'}[1+\tan(y)]+x\color{royalblue}{[1+\tan(y)]'}\]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?

  6. clara1223
    • one year ago
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    Oh! I had it written down but forgot to type it here.

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

  8. zepdrix
    • one year ago
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    \[\large\rm \frac{d}{dx}(1+\tan y)=?\]

  9. clara1223
    • one year ago
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    the 1 would become 0 and the tan y would be sec^2 (y) (dy/dx), is that right?

  10. zepdrix
    • one year ago
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    \[\large\rm \frac{d}{dx}(1+\tan y)=0+\sec^2(y)y'\]Ok great!

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

  12. zepdrix
    • one year ago
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    \[\large\rm \cos x=1+\tan y+xy' \sec^2 y\]So how do we do that? :) Any ideas?

  13. clara1223
    • one year ago
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    you lost me. did you just move the y' to the x in front of sec^2(y)?

  14. zepdrix
    • one year ago
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    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.

  15. clara1223
    • one year ago
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    before you did that I could isolate y' and get (cosx-tany-1)/(x(sec^2)y) on the other side

  16. clara1223
    • one year ago
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    We did a lot like that in class

  17. clara1223
    • one year ago
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    Is there something wrong with that?

  18. zepdrix
    • one year ago
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    \[\large\rm \cos x=1+\tan y+xy' \sec^2 y\]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:/

  19. zepdrix
    • one year ago
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    I'm not sure what (sec^2)y is. Careful with the way you do your brackets :)

  20. clara1223
    • one year ago
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    written down it looks identical to yours

  21. clara1223
    • one year ago
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    i'm just not too good at getting this stuff typed

  22. zepdrix
    • one year ago
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    oh i see :D

  23. zepdrix
    • one year ago
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    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.

  24. anonymous
    • one year ago
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    Refer to the attachment using Mathematica v9

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