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anonymous

  • one year ago

If the derivative of ​f[x_] = (x^2 - 2 Log[x]) ​f'[x_] = (2x - 2/x) Then why is the derivative of ​f[x_] = (x^2 - 2 Log[x])/2 ​f'[x_] = (2x - 2/x)/2 What is the principle that allows the /2 which appears to be some kind of constant to remain in the derivative ?

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  1. dan815
    • one year ago
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    because derative is a linear operator

  2. hartnn
    • one year ago
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    constants can be factored out of the derivative

  3. UsukiDoll
    • one year ago
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    wow this guy must be rich to do a QH question

  4. dan815
    • one year ago
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    ​f[x_] = 2*(x^2 - 2 Log[x]) heres another example this is same as ​f[x_] = (x^2 - 2 Log[x]) + (x^2 - 2 Log[x]) there are 2 of them, and if u remember we can differentiate each term separately

  5. anonymous
    • one year ago
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    no just desperate, lol .. I've given up caring about money

  6. anonymous
    • one year ago
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    so I just didnt simplify it enough?

  7. dan815
    • one year ago
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    no its not like that its just that the derivative is a linear operator

  8. UsukiDoll
    • one year ago
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    @plasmataco this is calculus ii material

  9. dan815
    • one year ago
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    you can see taking a derivative is a linear operator if you write out the first principles

  10. UsukiDoll
    • one year ago
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    A moderator is present in this conversation. Do not go off topic and break the rules

  11. dan815
    • one year ago
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    |dw:1434803107603:dw|

  12. UsukiDoll
    • one year ago
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    maybe we can try to take the derivatives together.. one is straight forward first it's just derivative and then second is a simple product rule .

  13. dan815
    • one year ago
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    |dw:1434803242178:dw|

  14. hartnn
    • one year ago
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    /2 is division by 2. multiplication or division of a constant to the function, multiplies or divides its derivative by the same constant.. \(\Large if ~ ~f'(x) = g(x), \\ then , ~~ [a \times f(x)]' =a \times g(x) \)

  15. anonymous
    • one year ago
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    ah ok.. that's the piece I was missing hartnn

  16. anonymous
    • one year ago
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    even though its a /2 .. it's a constant multiple.

  17. hartnn
    • one year ago
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    exactly!

  18. anonymous
    • one year ago
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    ah ok.. that makes things click for me.

  19. anonymous
    • one year ago
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    sorry dan.. that was a colossal effort, much appreciated.. you lost me a bit there though :)

  20. dan815
    • one year ago
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    thats all there is to it, i was showing you why, you can factor the constant out

  21. UsukiDoll
    • one year ago
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    that's why he's the honorary professor of Mathematics XD

  22. dan815
    • one year ago
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    lol :>

  23. anonymous
    • one year ago
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    I got to give this one to hartnn though.. that was the piece I was missing.. I knew the /2 had to carry into the derivative, but I could not work out for the life of me, what rule was making that happen.. once I realized it was just a 0.5 f[x] constant multiple, then it clicked..

  24. Plasmataco
    • one year ago
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    I gave mine to dan, so... yeah, give to @hartnn

  25. UsukiDoll
    • one year ago
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    mine is also given to dan

  26. hartnn
    • one year ago
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    thanks hugh! Its been years i didn't care about medals and stuff, so feel free to give it to anyone you like :)

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