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bmelyk Group Title

wondering if someone will walk me through how to do this problem:

  • one year ago
  • one year ago

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  1. bmelyk Group Title
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    Use the appropriate rule or combination of rules to find the derivative of the function defined below. \[y=\sec^2(t^2-1)\]

    • one year ago
  2. hartnn Group Title
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    first can u find the derivative of sec^2 x ?

    • one year ago
  3. bmelyk Group Title
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    sec^2tan^2

    • one year ago
  4. bmelyk Group Title
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    well to the power of 2x i guess.

    • one year ago
  5. hartnn Group Title
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    to the power of 2x ? u know chain rule, right ? what is derivative of x^2 first ?

    • one year ago
  6. bmelyk Group Title
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    2x

    • one year ago
  7. bmelyk Group Title
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    1/2rootx

    • one year ago
  8. hartnn Group Title
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    its 2x, right, so derivative of sec^2 x = ?

    • one year ago
  9. bmelyk Group Title
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    2x(sec)*(sec2xtan2x) ??

    • one year ago
  10. hartnn Group Title
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    applying chain rule, u get \((sec^2x)' = 2sec\: x(sec\: x)'=2sec \:x(sec\:xtan\:x)\) got this ?

    • one year ago
  11. bmelyk Group Title
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    yes i got it.

    • one year ago
  12. hartnn Group Title
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    so now can u apply chain rule again and find it for your question ?

    • one year ago
  13. bmelyk Group Title
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    i was just confused how you just had sec^2x wrote down is all, i thought you meant to the power of 2x, not sec^2(x)

    • one year ago
  14. hartnn Group Title
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    oh, sorry for confusion again....

    • one year ago
  15. bmelyk Group Title
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    haha its okay, theres just no x's present in the question is all.

    • one year ago
  16. bmelyk Group Title
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    so now i have: \[\sec^2(2t)+(t^2-1)(2)(\sec^2)(sectan)\]

    • one year ago
  17. hartnn Group Title
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    how sec^2 2t? the angle doesn't change...

    • one year ago
  18. bmelyk Group Title
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    im using the product rule.

    • one year ago
  19. hartnn Group Title
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    \([sec^2(t^2-1)]'= 2sec (t^2-1)[sec(t^2-1)]'\) got this step ?

    • one year ago
  20. bmelyk Group Title
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    yup/

    • one year ago
  21. hartnn Group Title
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    now next step would be ?

    • one year ago
  22. bmelyk Group Title
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    to use thechain rule on [sec(t2−1)

    • one year ago
  23. hartnn Group Title
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    yes. try it out.

    • one year ago
  24. bmelyk Group Title
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    sectan(t^2-1)(2t) ??

    • one year ago
  25. hartnn Group Title
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    yes! precisely it would be sec(t^2-1)tan(t^2-1)(2t)

    • one year ago
  26. hartnn Group Title
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    so final answe would be ?

    • one year ago
  27. bmelyk Group Title
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    so now i have: \[2\sec(t^2-1)*\sec(t^2-1)\tan(t^2-1)(2t)\]

    • one year ago
  28. hartnn Group Title
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    correct! than can be simplified to \(2\sec^2(t^2-1)*\tan(t^2-1)(2t)\)

    • one year ago
  29. hartnn Group Title
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    \(4t\sec^2(t^2-1)*\tan(t^2-1)\)

    • one year ago
  30. bmelyk Group Title
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    so that's my final answer? : )

    • one year ago
  31. hartnn Group Title
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    yup. any doubts ?

    • one year ago
  32. hartnn Group Title
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    but, did u get all steps ?

    • one year ago
  33. bmelyk Group Title
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    can i write that as: \[4t*\sec(t^2-1)^2*\tan(t^2-1)\]

    • one year ago
  34. bmelyk Group Title
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    i wrote them all down : ) i get them now, it's not as hard when i got someone explaining it to me hah.

    • one year ago
  35. bmelyk Group Title
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    can i write it like that though???

    • one year ago
  36. hartnn Group Title
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    yup.

    • one year ago
  37. bmelyk Group Title
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    should there be a *t at the end of the equation?

    • one year ago
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