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Calle87

  • 2 years ago

differentiate using the definition

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  1. Calle87
    • 2 years ago
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    \[\frac{ 1 }{ \sqrt{x} }\]

  2. bahrom7893
    • 2 years ago
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    it's lim as h->0 (f(x+h)-f(x))/h

  3. bahrom7893
    • 2 years ago
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    f(x+h) = 1/sqrt(x+h) f(x) = 1/sqrt(x)

  4. bahrom7893
    • 2 years ago
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    Do some algebra, cancel out the hs

  5. Calle87
    • 2 years ago
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    \[\frac{ \frac{ 1 }{ \sqrt{x+h} }-\frac{ 1 }{ \sqrt{x} } }{ h }\]

  6. bahrom7893
    • 2 years ago
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    yea, now multiply by the conjugate i think.. both top and bottom

  7. Calle87
    • 2 years ago
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    \[\frac{ \frac{ \sqrt{x}+\sqrt{x-h}}{ \sqrt{x}\sqrt{x+h} } }{ h }\]

  8. across
    • 2 years ago
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    Typo up there ^

  9. Calle87
    • 2 years ago
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    oops meant to put the h on the bottom :}

  10. bahrom7893
    • 2 years ago
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    it's supposed to be a minus

  11. bahrom7893
    • 2 years ago
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    all the way on top in the middle

  12. Calle87
    • 2 years ago
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    oh ya that to

  13. bahrom7893
    • 2 years ago
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    and the other one on top must be a plus

  14. across
    • 2 years ago
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    Inside the radical, too.

  15. bahrom7893
    • 2 years ago
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    u mixed up the signs

  16. Calle87
    • 2 years ago
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    oops!

  17. Calle87
    • 2 years ago
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    ok so i got all that part, the conjugates are what get me

  18. bahrom7893
    • 2 years ago
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    twiddla time then.. hang on a sec

  19. Calle87
    • 2 years ago
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    \[\frac{ -h }{ h \sqrt{x}\sqrt{x+h(\sqrt{x}+\sqrt{x+h})} }\]

  20. Calle87
    • 2 years ago
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    h goes away

  21. Calle87
    • 2 years ago
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    that square root is mess up in the bottom

  22. bahrom7893
    • 2 years ago
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    yup

  23. bahrom7893
    • 2 years ago
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    No, since h is 0, on the bottom u just have sqrt(x+0)

  24. bahrom7893
    • 2 years ago
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    so it's -1/sqrt(x) which is correct

  25. Calle87
    • 2 years ago
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    at what point do i make h = 0?

  26. bahrom7893
    • 2 years ago
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    when u dont run into trouble if u do.

  27. bahrom7893
    • 2 years ago
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    1/h <- NO 1/(4+1) <- YES

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