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NathanJHW

  • one year ago

If y^5 +3x^2y^2 +5x^4 = 49 , then dy/dx at the point (-1, 2) is:

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  1. NathanJHW
    • one year ago
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    @perl

  2. NathanJHW
    • one year ago
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    −1 11/23 −23/11 −10/3 0

  3. NathanJHW
    • one year ago
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    @Hero

  4. NathanJHW
    • one year ago
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    @zepdrix

  5. zepdrix
    • one year ago
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    \[\Large\rm y^5 +3x^2y^2 +5x^4 = 49\]So looks like you'll want to differentiate implicitly. Having trouble with that step? :o

  6. NathanJHW
    • one year ago
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    Yes I am.

  7. zepdrix
    • one year ago
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    We're taking our derivative with respect to x,\[\Large\rm (y^5)'\ne5y^4\]So whenever you differentiate a non-x thing, you end up with this dy/dx or y' term by the chain rule,\[\Large\rm (y^5)'=5y^4y'\]

  8. zepdrix
    • one year ago
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    \[\Large\rm \color{orangered}{(y^5)'} +(3x^2y^2)' +(5x^4)' = (49)'\]\[\Large\rm \color{orangered}{5y^4y'} +(3x^2y^2)' +(5x^4)' = (49)'\]The middle part is a little trickier. You have two variable things being multiplied together, so we'll need to apply our product rule. It doesn't matter which you group the 3 with, just choose one of them.\[\Large\rm 5y^4y' +\color{orangered}{(3x^2y^2)'} +(5x^4)' = (49)'\]\[\Large\rm 5y^4y' +\color{orangered}{(3x^2)'y^2+3x^2(y^2)'} +(5x^4)' = (49)'\]That product rule setup make sense? :o

  9. NathanJHW
    • one year ago
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    Yeah that makes sense to me.

  10. zepdrix
    • one year ago
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    So what do you get for the orange stuff? :o

  11. NathanJHW
    • one year ago
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    Do you mean what did I get when I did the work or do I solve your work now?

  12. zepdrix
    • one year ago
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    Oh you already attempt this problem? :) Yah either is fine, we can continue doing this, or you can tell me what you got doing your way and I can check it. either is fine

  13. NathanJHW
    • one year ago
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    We can continue with your work, I was no where close so it's all wrong.

  14. zepdrix
    • one year ago
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    So let's just worry about the orange part right now, what do you get when you take those derivatives? (In one of them, you're taking derivative of y, so don't forget about your dy/dx)

  15. NathanJHW
    • one year ago
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    ((dy/dx)(5y^4y)) +((dy/dx)(3x^2y^2))+ ((d/dx)(5x^4))=(d/dx)(49)

  16. NathanJHW
    • one year ago
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    Is that the correct setup @zepdrix

  17. zepdrix
    • one year ago
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    hmm

  18. zepdrix
    • one year ago
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    If you don't like the primes, fine we can use the differentials instead. So we have:\[\Large\rm \frac{d}{dx}(y^5 +3x^2y^2 +5x^4) = \frac{d}{dx}(49)\] \[\Large\rm \frac{d}{dx}(y^5) +\frac{d}{dx}(3x^2y^2) +\frac{d}{dx}(5x^4) = \frac{d}{dx}(49)\]First term gives you:\[\Large\rm \color{orangered}{5y^4\frac{dy}{dx}} +\frac{d}{dx}(3x^2y^2) +\frac{d}{dx}(5x^4) = \frac{d}{dx}(49)\]

  19. zepdrix
    • one year ago
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    I think you're having a little trouble with your chain rule... or something.. hmm

  20. zepdrix
    • one year ago
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    \[\Large\rm \frac{d}{dx}5x^4=5\cdot4x^3\cdot \frac{d}{dx}x=20x^3\cdot \frac{dx}{dx}=20x^3\cdot 1\]With your third term here, keep in mind that we're differentiating with respect to x, so you don't end up with a dy/dx on the last term, you actually end up with a "dx/dx" which is insignificant.

  21. zepdrix
    • one year ago
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    Alternatively what I did with the y term was,\[\Large\rm \frac{d}{dx}y^5=5y^4\cdot\frac{d}{dx}y=5y^4\cdot \frac{dy}{dx}\]chain rule again, but now we end up with something that actually has meaning, this dy/dx.

  22. zepdrix
    • one year ago
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    |dw:1433291724073:dw|You're always applying your chain rule, even just with the x's. We often don't show that work because it has no significance. If we're differentiating x with respect to x, we don't end up with any extra stuff.

  23. zepdrix
    • one year ago
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    Too confusing? :O

  24. NathanJHW
    • one year ago
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    So how does this lead into the answer?

  25. NathanJHW
    • one year ago
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    I understand how to get this, but where do we go next?

  26. zepdrix
    • one year ago
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    Well we've gotta get these terms set up correctly. Your middle derivative is looking strange.

  27. zepdrix
    • one year ago
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    First term gave us this,\[\Large\rm \color{orangered}{5y^4\frac{dy}{dx}} +\frac{d}{dx}(3x^2y^2) +\frac{d}{dx}(5x^4) = \frac{d}{dx}(49)\]Third term will give us this, no dy/dx on this guy,\[\Large\rm 5y^4\frac{dy}{dx} +\frac{d}{dx}(3x^2y^2) +\color{orangered}{5\cdot4x^3} = \frac{d}{dx}(49)\]

  28. zepdrix
    • one year ago
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    Right side will give us this:\[\Large\rm 5y^4\frac{dy}{dx} +\frac{d}{dx}(3x^2y^2) +5\cdot4x^3 = \color{orangered}{0}\]

  29. zepdrix
    • one year ago
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    Middle term we have to setup our product rule before we apply any differentiation rules like power rule or chain rule. \[\large\rm 5y^4\frac{dy}{dx} +\color{royalblue}{y^2\frac{d}{dx}(3x^2)+3x^2\frac{d}{dx}(y^2)} +5\cdot4x^3 = 0\]

  30. zepdrix
    • one year ago
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    And then apply your rules to each term individually, you shouldn't end up with a dy/dx on the whole group, i was a little confused by that.

  31. zepdrix
    • one year ago
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    \[\large\rm 5y^4\frac{dy}{dx} +\color{orangered}{y^2(6x)}+\color{royalblue}{3x^2\frac{d}{dx}(y^2)} +5\cdot4x^3 = 0\]

  32. zepdrix
    • one year ago
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    And the other part gives you,\[\large\rm 5y^4\frac{dy}{dx} +y^2(6x)+\color{orangered}{3x^2(2y)\frac{dy}{dx}} +5\cdot4x^3 = 0\]

  33. zepdrix
    • one year ago
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    And then you want to evaluate your derivative function dy/dx at the point (-1,2). So plug in `-1 for all the x` and `2 for all the y`.

  34. zepdrix
    • one year ago
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    \[\large\rm 5(2)^4\frac{dy}{dx} +(2)^2(6(-1))+3(-1)^2(2(2))\frac{dy}{dx} +5\cdot4(-1)^3 = 0\]Something like that, ya? I probably used more brackets than I needed to -_- Anyway, then you solve for dy/dx.

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