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bmelyk

  • 2 years ago

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

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

  2. hartnn
    • 2 years ago
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    first can u find the derivative of sec^2 x ?

  3. bmelyk
    • 2 years ago
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    sec^2tan^2

  4. bmelyk
    • 2 years ago
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    well to the power of 2x i guess.

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

  6. bmelyk
    • 2 years ago
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    2x

  7. bmelyk
    • 2 years ago
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    1/2rootx

  8. hartnn
    • 2 years ago
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    its 2x, right, so derivative of sec^2 x = ?

  9. bmelyk
    • 2 years ago
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    2x(sec)*(sec2xtan2x) ??

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

  11. bmelyk
    • 2 years ago
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    yes i got it.

  12. hartnn
    • 2 years ago
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    so now can u apply chain rule again and find it for your question ?

  13. bmelyk
    • 2 years ago
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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)

  14. hartnn
    • 2 years ago
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    oh, sorry for confusion again....

  15. bmelyk
    • 2 years ago
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    haha its okay, theres just no x's present in the question is all.

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

  17. hartnn
    • 2 years ago
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    how sec^2 2t? the angle doesn't change...

  18. bmelyk
    • 2 years ago
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    im using the product rule.

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

  20. bmelyk
    • 2 years ago
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    yup/

  21. hartnn
    • 2 years ago
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    now next step would be ?

  22. bmelyk
    • 2 years ago
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    to use thechain rule on [sec(t2−1)

  23. hartnn
    • 2 years ago
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    yes. try it out.

  24. bmelyk
    • 2 years ago
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    sectan(t^2-1)(2t) ??

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

  26. hartnn
    • 2 years ago
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    so final answe would be ?

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

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

  29. hartnn
    • 2 years ago
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    \(4t\sec^2(t^2-1)*\tan(t^2-1)\)

  30. bmelyk
    • 2 years ago
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    so that's my final answer? : )

  31. hartnn
    • 2 years ago
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    yup. any doubts ?

  32. hartnn
    • 2 years ago
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    but, did u get all steps ?

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

  34. bmelyk
    • 2 years ago
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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.

  35. bmelyk
    • 2 years ago
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    can i write it like that though???

  36. hartnn
    • 2 years ago
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    yup.

  37. bmelyk
    • 2 years ago
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    should there be a *t at the end of the equation?

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