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clara1223

  • one year ago

find f'(x) of f(x)=cos(1/x)+sec(8x)

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  1. zepdrix
    • one year ago
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    Ooo trig! :O Fun stuff!

  2. zepdrix
    • one year ago
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    Do you know derivative of sec(x)? :)

  3. clara1223
    • one year ago
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    yes, tan(x)sec(x)

  4. zepdrix
    • one year ago
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    \[\large\rm \frac{d}{dx}\sec(x)=\sec(x)\tan(x)\]\[\large\rm \frac{d}{dx}\sec(8x)=\sec(8x)\tan(8x)\cdot\color{royalblue}{(8x)'}\]Good good good. So to finish up the derivative of the second term, you need only make sure you apply your chain rule.

  5. clara1223
    • one year ago
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    so the derivative of sec(8x) is sec(8x)tan(8x)?

  6. zepdrix
    • one year ago
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    Not entirely :O That's most of it. Do you see the blue portion though? You have to multiply by the derivative of the `inner function`.

  7. zepdrix
    • one year ago
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    Derivative of 8x? :o

  8. clara1223
    • one year ago
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    8

  9. zepdrix
    • one year ago
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    \[\large\rm \frac{d}{dx}\sec(8x)=\sec(8x)\tan(8x)\cdot\color{royalblue}{(8x)'}=8\sec(8x)\tan(8x)\] Ok good! :) That takes care of the second term.

  10. zepdrix
    • one year ago
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    An extra 8 popping out due to our chain rule.

  11. xapproachesinfinity
    • one year ago
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    chain rule, whenever stuff are inside think of performing that

  12. zepdrix
    • one year ago
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    Remember your cosine derivative? :)

  13. clara1223
    • one year ago
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    -sin(x)

  14. xapproachesinfinity
    • one year ago
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    i mean stuff except that not of the form x only

  15. zepdrix
    • one year ago
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    \[\large\rm \frac{d}{dx}\cos(x)=-\sin(x)\]Good. We'll follow the same pattern, applying chain rule again though.\[\large\rm \frac{d}{dx}\cos\left(\frac{1}{x}\right)=-\sin\left(\frac{1}{x}\right)\cdot\color{royalblue}{\left(\frac{1}{x}\right)'}\]

  16. zepdrix
    • one year ago
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    Do you remember how to differentiate something when it's in the denominator like that? :)

  17. clara1223
    • one year ago
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    (1/x)' = (x^-1)' = (-x^-2) = (-1/x^2)

  18. clara1223
    • one year ago
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    correct?

  19. xapproachesinfinity
    • one year ago
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    yes good!

  20. zepdrix
    • one year ago
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    \[\large\rm \frac{d}{dx}\cos\left(\frac{1}{x}\right)=-\sin\left(\frac{1}{x}\right)\cdot\color{orangered}{\left(-\frac{1}{x^2}\right)}\]Yayyy good job \c:/ Maybe simplify things down a tad, bring the negatives together and such.\[\large\rm \frac{d}{dx}\cos\left(\frac{1}{x}\right)=\frac{1}{x^2}\cdot\sin\left(\frac{1}{x}\right)\]

  21. clara1223
    • one year ago
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    Does that need to be simplified any further?

  22. zepdrix
    • one year ago
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    No, not unless you wanted to get a common denominator between the two terms. That seems unnecessary though.

  23. anonymous
    • one year ago
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    Just thought I would add the trick to remember the derivative to those nastier trig functions: sec(x)=1/cos(x): \[cos^{-1}(x)\] and therefore by the chain rule: \[\frac{d}{dx}cos^{-1}(x)=-1*cos^{-2}(x)*(-sin(x)=tan(x)*cos^{-1}(x)=tan(x)sec(x)\]

  24. clara1223
    • one year ago
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    so the final answer is \[\frac{ 1 }{ x ^{2} }\sin(\frac{ 1 }{ x })+8\sec(8x)\tan(8x)\] ?

  25. anonymous
    • one year ago
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    you can do the same with cosec(x), tan(x), etc

  26. zepdrix
    • one year ago
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    yay good job \c:/

  27. xapproachesinfinity
    • one year ago
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    oh no there a down side on writing 1/cos that way you will get mixed up with inverse function

  28. xapproachesinfinity
    • one year ago
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    just leave it as it is!

  29. xapproachesinfinity
    • one year ago
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    or use parenthesis (cosx)^-1

  30. anonymous
    • one year ago
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    Oh yeah sorry.... implied by the context but your right poor form for those not as comfortable im sorry :/

  31. anonymous
    • one year ago
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    I usually use arc... but I get your point

  32. xapproachesinfinity
    • one year ago
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    yeah but now everyone use arc lol we use them interchangeably

  33. xapproachesinfinity
    • one year ago
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    not* everyone

  34. xapproachesinfinity
    • one year ago
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    check this derivative carla f(x)=tan(cosx)

  35. xapproachesinfinity
    • one year ago
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    see if you mastered the rule lol

  36. xapproachesinfinity
    • one year ago
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    @clara1223

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