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calyne

Differentiate the function y = ln |2-x-5x^2|

  • 2 years ago
  • 2 years ago

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  1. DHASHNI
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    is the ques correct?

    • 2 years ago
  2. calyne
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    yup

    • 2 years ago
  3. calyne
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    oops

    • 2 years ago
  4. calyne
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    fixed it

    • 2 years ago
  5. calyne
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    ...

    • 2 years ago
  6. amistre64
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    \[[ln(u)]'=\frac{u'}{u}\]

    • 2 years ago
  7. calyne
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    what are those brackets or ||

    • 2 years ago
  8. calyne
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    because it's not |lnu| it's ln|u|

    • 2 years ago
  9. amistre64
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    those are just for grouping; it means that the whole of it is derived and not just the (u) part

    • 2 years ago
  10. calyne
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    okay so show me how

    • 2 years ago
  11. amistre64
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    i believe it goes piece wise in the end

    • 2 years ago
  12. amistre64
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    or you can use the sqrt(u)^2 trick

    • 2 years ago
  13. calyne
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    please show me

    • 2 years ago
  14. DHASHNI
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    |dw:1334421083017:dw|

    • 2 years ago
  15. amistre64
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    \[\frac{d}{dx}ln(\sqrt{u\ }^2)=\frac{(\sqrt{u\ }^2)'}{ \sqrt{u\ }^2}\to\ \frac{2\sqrt{u\ }*\sqrt{u}'}{ \sqrt{u\ }^2}\]

    • 2 years ago
  16. amistre64
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    if i follow that derivatives i think it would look like this ... but best to check with the wolf

    • 2 years ago
  17. amistre64
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    http://www.wolframalpha.com/input/?i=derivative+ln+%7C2-x-5x%5E2%7C the wolf doesnt seem to go thru the hoopla tho

    • 2 years ago
  18. amistre64
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    \[\frac{2\sqrt{u\ }}{ \sqrt{u\ }^2}\sqrt{u}'\] \[\frac{2\sqrt{u\ }}{ \sqrt{u\ }^2}\frac{u'}{2\sqrt{u}}=\frac{u'}{ \sqrt{u\ }^2}\to\ \frac{u'}{ |u|}\]

    • 2 years ago
  19. calyne
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    whatevr it's 1/(2-x-5x^2) * -(10x+1) = 10x+1 / 5x^2+x-2 woohoo

    • 2 years ago
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