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calyne

  • 2 years ago

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

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  1. DHASHNI
    • 2 years ago
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    is the ques correct?

  2. calyne
    • 2 years ago
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    yup

  3. calyne
    • 2 years ago
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    oops

  4. calyne
    • 2 years ago
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    fixed it

  5. calyne
    • 2 years ago
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    ...

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

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

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

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

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

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

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

  13. calyne
    • 2 years ago
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    please show me

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

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

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

  17. amistre64
    • 2 years ago
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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

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

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

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