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anonymous

  • 5 years ago

how do i find the derivative of: Y=x^3 tan^2 x

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  1. myininaya
    • 5 years ago
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    you have a function times a function so you have to use the product rule

  2. myininaya
    • 5 years ago
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    so we have y'=[x^3]'(tanx)^2+x^3[(tanx)^2]'

  3. myininaya
    • 5 years ago
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    (x^3)'=3x^2 using power rule

  4. myininaya
    • 5 years ago
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    we have [(tanx)^2]'=2(tanx)*[secx]^2 using chain rule

  5. anonymous
    • 5 years ago
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    thank you so much

  6. myininaya
    • 5 years ago
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    np

  7. anonymous
    • 5 years ago
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    \[3x^2(\tan^2x)+x^3(1/\cos^2x)\]

  8. anonymous
    • 5 years ago
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    thank you

  9. anonymous
    • 5 years ago
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    anytime :)

  10. myininaya
    • 5 years ago
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    That's not right mmonish91

  11. anonymous
    • 5 years ago
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    we know the equation is uv' + u'v, so take the following: - u = x^3 - u' = 3x^2 - v = tan^2x - v' = 2 tanx sec^2x so: y' = uv' + u'v = (x^3)(2tanxsec^2x) + (3x^2)(tan^2x)

  12. anonymous
    • 5 years ago
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    ^_^ simply as that.

  13. anonymous
    • 5 years ago
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    simple*

  14. anonymous
    • 5 years ago
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    @myinninaya please do explain

  15. anonymous
    • 5 years ago
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    mmonish dear, you started correctly, but didn't get d/dx tan^2x right. the derivative of tanx = sec^2x , but in this case the derivative of: (tanx)^ 2 = 2.(tanx)(sec^2x) ; you multiply the power first then subtract it, then you derive tanx. (power rule) ^_^ a^ u = u(a)(u') <-- rule :)

  16. anonymous
    • 5 years ago
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    i see...i left out the 2 for tan^2 mah bad

  17. anonymous
    • 5 years ago
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    a^ u = u(a)^u-1 (u') , sorry lol

  18. anonymous
    • 5 years ago
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    no problem, we all make mistakes ^_^

  19. anonymous
    • 5 years ago
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    i don't really work with sec so my professor said we could leave it in sin or cos format

  20. anonymous
    • 5 years ago
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    but you must know the derivative of tanx = sec^2x while the derivative of secx = secxtanx :)

  21. anonymous
    • 5 years ago
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    which is same as 1/cos^2x

  22. anonymous
    • 5 years ago
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    but I see what you mean :), u left it in cos, it's alright

  23. anonymous
    • 5 years ago
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    yes, true :)

  24. anonymous
    • 5 years ago
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    thank you guys!! ^.^

  25. anonymous
    • 5 years ago
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    you are most welcome sindy ^_^ glad you understood it

  26. anonymous
    • 5 years ago
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    yes i did understand :D ..

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