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marigirl

  • one year ago

Implicit differentiation: I understand that you would need to use it for cases where the question is to differentiate y=x^x But how do i know when to use it? Could you see the picture i attached below, implicit differentiation was used, however, i could have made it the subject and gone about differentiating it explicitly..

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  1. marigirl
    • one year ago
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    When i did this question, i made r the subject by taking the square root of the left side

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  2. zepdrix
    • one year ago
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    When to use it? Hmm

  3. zepdrix
    • one year ago
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    You use it when isolating y is more work than it's worth, sometimes it's even impossible

  4. zepdrix
    • one year ago
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    But for your problem, you're saying that end up with a square root which seems different from the correct answer or something?

  5. marigirl
    • one year ago
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    okay, and yes in my question What i did was i did \[r=\sqrt{60h-h^2}\] and i thought that was not too bad to differentiate......but the model answers went about implicit differentiating it

  6. zepdrix
    • one year ago
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    Well you get something that is `not a function` when you solve for r directly,\[\large r=\pm\sqrt{60h-h^2}\]Don't forget the +/- when you take the root of a square! :) Maybe that's why they avoided going that way, hmm.

  7. zepdrix
    • one year ago
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    So doing it implicitly, you end up with:\[\large\rm r'=\frac{30-h}{r}\]If we plug r into this equation:\[\large\rm r'=\pm\frac{30-h}{\sqrt{30^2-(30-h)^2}}\]It probably looks more or less the same as what you got, going the other route, yes? :o

  8. marigirl
    • one year ago
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    yes ..so i worried about nothing

  9. zepdrix
    • one year ago
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    XD

  10. marigirl
    • one year ago
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    Thanks :) I appreciate it

  11. zepdrix
    • one year ago
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    This MIT video gives an example of this type of thing happening https://youtu.be/5q_3FDOkVRQ?t=9m10s Just in case you wanted to have a better understanding of what is going on, he can explain it much better than I can :D lol

  12. marigirl
    • one year ago
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    I wish to be confident like you in maths one day :) your truly an inspiration

  13. zepdrix
    • one year ago
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    aw :3 lol

  14. marigirl
    • one year ago
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    hey

  15. marigirl
    • one year ago
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    \[y=x^{\frac{ m }{ n }}\] how did that become \[y^n=x^m\]

  16. zepdrix
    • one year ago
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    Start with a simple example to convince yourself:\[\large\rm y=\sqrt{x}\]If I rewrite this as:\[\large\rm y=x^{1/2}\]In either form, I can see that I need to square to get rid of the root,\[\large\rm y^2=x\]This will happen with any degree of root, your exponent is always the reciprocal of the power on x.\[\large\rm y=\sqrt[3]{x}\qquad\to\qquad y=x^{1/3}\qquad\to\qquad y^3=x\]Err I guess I should be careful the way I say reciprocal, because we really don't care about the numerator on x. So back to the example you gave:\[\large\rm y=x^{m/n}\]To deal with this n'th root, I'm raising both sides to the n'th power,\[\large\rm y^n=\left(x^{m/n}\right)^n\]Rules of exponents gives:\[\large\rm y^n=x^{\frac{m}{n}\cdot n}\]\[\large\rm y^n=x^m\]

  17. marigirl
    • one year ago
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    yes i see it.. thank you so much. im off to watch more MIT videos now :D

  18. zepdrix
    • one year ago
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    c:

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