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xcrypt

  • 2 years ago

This is probably a weird question but is there a reason or intuition we write higher order derivatives as \[ \frac{d^n y}{dx^n} \] instead of \[ \frac{d^n y}{d^n x} \] or something similar?

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  1. thinker1
    • 2 years ago
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    no man|dw:1352457865514:dw|

  2. thinker1
    • 2 years ago
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    do u understand mate

  3. xcrypt
    • 2 years ago
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    that's not what I am asking though

  4. lgbasallote
    • 2 years ago
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    do you agree that \[\huge \frac{dy}{dx} = \frac d{dx}\] they're the same. right?

  5. xcrypt
    • 2 years ago
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    I have never the right hand side notation before

  6. lgbasallote
    • 2 years ago
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    you're not familiar with d/dx?

  7. xcrypt
    • 2 years ago
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    no

  8. xcrypt
    • 2 years ago
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    is it the same by definition?

  9. lgbasallote
    • 2 years ago
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    that's a bummer...i would assume you're at the start of your calculus class...hmm i'll try thinking of another analogy

  10. xcrypt
    • 2 years ago
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    I'm actually doing real analysis

  11. lgbasallote
    • 2 years ago
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    ...then why are you not familiar with d/dx?

  12. xcrypt
    • 2 years ago
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    I have never seen it before, not in my books

  13. xcrypt
    • 2 years ago
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    But I've never taken calculus before ra so that might explain it

  14. lgbasallote
    • 2 years ago
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    hmm i can't think of any other way to explain the reason behind d^n y/dx^n without using d/dx...so i suppose i should explain that

  15. thinker1
    • 2 years ago
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    actually i wanted to show u that if \[d ^{n}y /d ^{n} x\]would have meant that its reciprocal had just an inverse relationship with it but actual \[d ^{n}y /d ^{n} x\] is differently related to \[d ^{n}x /d ^{n} y\]

  16. lgbasallote
    • 2 years ago
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    \[\frac{dy}{dx} = \frac d{dx}\] mainly because \[\frac{dy}{dx} \implies \frac{d}{dx} (y)\] (this is algebra) then since y is usuallyimplied in functions, it is usually ommitted..that's why they just write d/dx for dy/dx

  17. UnkleRhaukus
    • 2 years ago
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    the derivative of a function is the rate that dependent variable changes as the independent variable varies in order

  18. lgbasallote
    • 2 years ago
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    then... let's say you want the second derivative...the second derivative would be \[\frac d{dx} (\frac{dy}{dx})\] so again...using algebra, you get \[\frac{d^2y}{dx^2}\] i don't know if that made anything clearer for you....

  19. xcrypt
    • 2 years ago
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    oh, we write f' or dy/dx instead of d/dx

  20. lgbasallote
    • 2 years ago
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    d/dx means derivative of with respect to x so dy/dx means derivative of y with respect to x

  21. xcrypt
    • 2 years ago
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    ah ok

  22. UnkleRhaukus
    • 2 years ago
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    \[y=x^2\]\[\frac{\text dy}{\text dx}=2x\]\[\frac{\text d^2y}{\text dx^2}=2\] \[\frac{\text d^2y}{\text dx}=2\text dx\]\[{\text d^2y}=2\text dx{\text dx}\]\[{\iint \text d^2y}=2\iint\text dx^2\]\[\int \text dy=2\int( x+c)\text dx\]\[y=2\frac{x^2}2+c\]\[y={x^2}+cx+d\]

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