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anonymous

  • 2 years ago

find the derivative of the function. arc sec2x

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  1. anonymous
    • 2 years ago
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    what do you have so far

  2. zepdrix
    • 2 years ago
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    Angela, are you allowed to use a table for derivatives of inverse trig functions? If not, there is a way we can do this using a triangle, it just takes a few more steps than normal.

  3. anonymous
    • 2 years ago
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    I honestly don't have a clue of wat to do :( @Decart @zepdrix

  4. zepdrix
    • 2 years ago
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    Answer the question silly billy >:O Are you given access to a table of Inverse Trig Derivatives? :)

  5. anonymous
    • 2 years ago
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    no I don't think so:) @zepdrix

  6. zepdrix
    • 2 years ago
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    \[\Large\bf y=arcsec 2x\qquad\qquad\to\qquad\qquad 2x=\sec y\]Understand what I did so far? I just rewrote the relationship so we don't have that inverse function anymore.

  7. anonymous
    • 2 years ago
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    that's possible?? :o and yes I understand:)

  8. zepdrix
    • 2 years ago
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    So what we want to do is some sneaky triangle math that you might remember back from Trig. \[\Large\bf \sec y\quad=\quad 2x\quad=\quad \frac{2x}{1}\quad=\quad \frac{hypotenuse}{adjacent}\]We want to draw this relationship on a triangle.

  9. zepdrix
    • 2 years ago
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    |dw:1384737357099:dw|Understand how we'll label this?

  10. anonymous
    • 2 years ago
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    lol let me look at it a lil more xp

  11. anonymous
    • 2 years ago
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    wait, does this apply if its f(x)=arc sec2x ??:) @zepdrix

  12. zepdrix
    • 2 years ago
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    You'll see how it connects later on D:

  13. anonymous
    • 2 years ago
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    ok lol go on:) @zepdrix

  14. zepdrix
    • 2 years ago
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    |dw:1384737660359:dw|So we'll label our sides according to our trig identity.

  15. zepdrix
    • 2 years ago
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    We can find the missing side using the Pythagorean Theorem, yes?

  16. anonymous
    • 2 years ago
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    yes:)

  17. zepdrix
    • 2 years ago
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    |dw:1384737872951:dw|

  18. zepdrix
    • 2 years ago
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    So we've successfully setup a triangle that relates back to secant.

  19. zepdrix
    • 2 years ago
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    NOW let's take a derivative.\[\Large\bf \sec y\quad=\quad 2x\]Taking our derivative with respect to x, what do you get? :x

  20. anonymous
    • 2 years ago
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    opposite side= sqrt of 4x^2-1 ?

  21. anonymous
    • 2 years ago
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    2 ?? @zepdrix

  22. zepdrix
    • 2 years ago
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    2 ?? water you talking about? :u

  23. anonymous
    • 2 years ago
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    the derivative ?? @zepdrix

  24. zepdrix
    • 2 years ago
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    that gives us the derivative of the `right side`, yes :) 2x turns into 2. What about the left side?

  25. anonymous
    • 2 years ago
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    sec y tan y ??@zepdrix

  26. zepdrix
    • 2 years ago
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    Mmm close!! Don't forget your chain rule! We're taking the derivative of a y-thing with respect to x. So the chain rule should also give us a y'

  27. zepdrix
    • 2 years ago
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    \[\Large \bf \sec y \tan y \cdot y'\quad=\quad 2\]

  28. anonymous
    • 2 years ago
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    oooh! @zepdrix

  29. zepdrix
    • 2 years ago
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    \[\huge \bf \sec y \tan y \cdot \color{#DD4747 }{y'}\quad=\quad 2\]Remember what this y represented at the start of the problem? \(\Large y=\sec 2x\) So this y' here represents the derivative of sec2x, the thing we've been looking for all along! :O

  30. zepdrix
    • 2 years ago
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    So we just need to isolate the pretty red y'

  31. zepdrix
    • 2 years ago
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    and write it in terms of x* which we can do using the triangle.

  32. zepdrix
    • 2 years ago
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    So if we divide some stuff over:\[\Large \bf y'\quad=\quad \frac{2}{\sec y \tan y}\]Then we'll recall our identities to rewrite this as:\[\Large\bf y'\quad=\quad 2\cos y \cot y\]

  33. zepdrix
    • 2 years ago
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    Then we use our triangle to get our final answer:\[\Large\bf y'\quad=\quad 2\frac{adjacent}{hypotenuse}\cdot \frac{adjacent}{opposite}\]

  34. anonymous
    • 2 years ago
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    and that's it?? @zepdrix

  35. zepdrix
    • 2 years ago
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    So when you plug those sides in you'll get a big ole mess that you want to simplify :o

  36. zepdrix
    • 2 years ago
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    Err a little mess :)

  37. zepdrix
    • 2 years ago
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    Plugging in the sides gives us: \[\Large\bf y'\quad=\quad 2\left(\frac{1}{2x}\right)\left(\frac{1}{\sqrt{4x^2-1}}\right)\] Mmm something like that.

  38. zepdrix
    • 2 years ago
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    If that process was wayyyy confusing, maybe just use a table of derivatives for now:\[\Large\bf y=\sec x\qquad\qquad\to\qquad\qquad y'\quad=\quad \frac{1}{x \sqrt{x^2-1}}\]

  39. zepdrix
    • 2 years ago
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    Since we had 2x,\[\Large\bf y=\sec \color{#DD4747}{2x}\qquad\to\qquad y'\quad=\quad \frac{1}{\color{#DD4747}{2x} \sqrt{(\color{#DD4747}{2x})^2-1}}(\color{#DD4747}{2x})'\]

  40. anonymous
    • 2 years ago
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    and just solve?? @zepdrix

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