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amyna

  • one year ago

y=-3(sec^2x - tan^2(x)) + 4x^3 I do not know how to solve this! Please help with the calc problem!! Thank you!

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  1. dan815
    • one year ago
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    sec(x) = 1/cos(x)

  2. freckles
    • one year ago
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    recall Pythagorean identities \[\sin^2(x)+\cos^2(x)=1 \\ \text{ divide both sides by } \cos^2(x) \\ \tan^2(x)+1=\sec^2(x) \\\] anyways I don't know what it is meant by solve

  3. freckles
    • one year ago
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    could you say what solve means here @amyna if you are here

  4. freckles
    • one year ago
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    you say this is a cal problem do you really mean differentiate are you trying to find the inverse like what are you doing exactly

  5. amyna
    • one year ago
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    yup sorry! differentiate it, thanks!

  6. freckles
    • one year ago
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    so you need help to differentiate -3+4x^3 just use constant rule and constant multiple rule and power rule?

  7. freckles
    • one year ago
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    -3 is a constant the derivative of a constant is 0

  8. freckles
    • one year ago
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    \[\frac{d}{dx} cx^n \\ c \frac{d}{dx} x^n \text{ by constant multiple rule } \\ c n x^{n-1} \text{ by power rule }\]

  9. freckles
    • one year ago
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    \[\frac{d}{dx}(-3)=? \\ \frac{d}{dx}4x^3=?\]

  10. amyna
    • one year ago
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    |dw:1436817831485:dw|

  11. freckles
    • one year ago
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    where did all the sec and tan stuff come from

  12. freckles
    • one year ago
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    did you use the Pythagorean identity earlier ? you should y=-3+4x^3

  13. freckles
    • one year ago
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    and we are wanting to find y' now

  14. freckles
    • one year ago
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    y'=0+12x^2=12x^2

  15. freckles
    • one year ago
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    though if you really want to differentiate sec^2(x)-tan^2(x) which is a constant so the derivative should be 0 but pretending you don't know you should wind up with this: \[\frac{d}{dx} \sec^2(x)=2 \sec(x) \cdot \sec(x)\tan(x) =2\sec^2(x)\tan(x) \\ \frac{d}{dx} \tan^2(x)=2 \tan(x) \sec^2(x) =2 \sec^2(x)\tan(x) \\ \text{ but as you notice } \\ \frac{d}{dx} \sec^2(x)-\frac{d}{dx} \tan^2(x)=0 \\ \frac{d}{dx}(\sec^2(x)-\tan^2(x))=0\] this is because \[\sec^2(x)-\tan^2(x) \text{ is a constant } \\ \text{ this is because } \sec^2(x)-\tan^2(x)=1 \text{ for all } x \text{ on it's domain }\]

  16. freckles
    • one year ago
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    but as I said all of that work is unnecessary

  17. freckles
    • one year ago
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    your y really just say y=-3+4x^3

  18. amyna
    • one year ago
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    so what is the final answer?

  19. freckles
    • one year ago
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    do you know how to differentiate -3 and 4x^3?

  20. amyna
    • one year ago
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    yes its 0 and 12x^2

  21. freckles
    • one year ago
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    ok great then why are you asking me what y' is when you know \[y=-3+4x^3 \\ y'=0+12x^2 \\ y'=12x^2 \]

  22. amyna
    • one year ago
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    oh okay! i was looking at it wrong! now i get it! thank you so much!

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