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jayshane

how about differentiate: y=sec²x-tan²x

  • one year ago
  • one year ago

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  1. terenzreignz
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    Derivative of the sum is the sum of the derivatives.

    • one year ago
  2. terenzreignz
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    Although, to be really creative, you can just recall that the derivative of a constant is 0 :D

    • one year ago
  3. terenzreignz
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    sec²x = tan²x + 1

    • one year ago
  4. wio
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    \[ \sec^2x-\tan^2x = \frac{1}{\cos^2x} - \frac{\sin^2 x}{\cos^2 x} \]

    • one year ago
  5. wio
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    So I mean you're gonna end up with y=1

    • one year ago
  6. kirbykirby
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    \[\frac{d}{dx}(\sec^2x-\tan^2x)=2\sec x(\sec x \tan x)-2\tan x(\sec^2x) = 2\sec^2x tanx - 2\sec^2x \tan x = 0 \]

    • one year ago
  7. kirbykirby
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    = 0 at the end

    • one year ago
  8. jayshane
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    the answer in the book is zero...

    • one year ago
  9. tkhunny
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    Please listen to @terenzreignz y = 1 Don't miss these sipmle ones.

    • one year ago
  10. kirbykirby
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    Sorry it cut off a but, but the last one just cut off "tan x"

    • one year ago
  11. terenzreignz
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    So many different ways to do it, but if all else eludes you, do it directly, as @kirbykirby But sec²x = tan²x + 1 So sec²x - tan²x = tan²x + 1 - tan²x = 1 And the derivative of 1...?

    • one year ago
  12. kirbykirby
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    Yes there are many ways to approach this one. You can re-write sec^x and tan^x in terms of cos x and sin x, but if you know the derivative of sec x and tan x (I recommend that you t do), then it's easier to proceed that way

    • one year ago
  13. jayshane
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    i will get the double prime if i will get the derivative 1...?

    • one year ago
  14. terenzreignz
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    It might pay (and save you time) to know by heart your trigonometric identities, too.

    • one year ago
  15. tkhunny
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    It absolutely WILL pay huge dividends to brush up on your trigonometry. If you were taking my examine and you did ANYTHING on this problem except recognize IMMEDIATELY that \(\dfrac{d}{dt}[\sec^{2}(t) - \tan^{2}(t)\ = \dfrac{d}{dt}1 = 0\), I would consider recommending you for a refresher course in a lower level class. Gotta know your trig. It's not optional!

    • one year ago
  16. kirbykirby
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    @jayshane I'm not sure what you mean about getting a double prime? But the derivative of 1 (or any constant) is just 0 :)

    • one year ago
  17. kirbykirby
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    Ok don't get so uptight @tkhunny :S Oye. It's not THE end of the world to not see it immediately o_o. I;m past cal 3 and doing differential equations now and even I didn't see it right away as you mentioned it lool. And I mean I barely use trig anymore that I am doing stats now

    • one year ago
  18. kirbykirby
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    I usually only think of these identities when I can't "plug and chug" right away lol

    • one year ago
  19. tkhunny
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    @kirbykirby No uptight going on here, but I do appreciate your calming influence. I feel very strongly about this particular issue. One should not be in differential equations without a very solid foundation in trigonometry. Simple as that. Also, as we have yet to see ANY effort by @jayshane, it's just not clear what we're doing, here. A little strong reality therapy seemed like it might be appropriate.

    • one year ago
  20. kirbykirby
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    Well I don't think that's a fair assumption to make that I don't have a good foundation in trig because I didn't recognize that as THE first thing to do?! o_o

    • one year ago
  21. jayshane
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    i acctually having a hard time dealing with trigonometric identities.. and im studying harder and harder on it its just im just an ordinary man.. but im really tring hard pls dont be anger on me...

    • one year ago
  22. terenzreignz
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    Nobody's angry with you @jayshane But, if you have the time, do study trigonometry... :)

    • one year ago
  23. kirbykirby
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    No we are not angry at you :) We were just recommending to really know the trig identities as much as you can because they save you time :)

    • one year ago
  24. jayshane
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    ive been memorizing it but i dont know how to use them in solving..

    • one year ago
  25. tkhunny
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    @ jayshane No anger, just encouragement. Get a trig book and work all 4,000 problems. Get it on your head and it will pay great dividends. @kirbykirby seems to be a good case that it isn't absolutely necessary, but if you could spend ALL your time in Differential Equations studying Differential Equations, don't you think this would be better than spending 1/2 your time in Differential Equations trying to remember trigonometry?

    • one year ago
  26. jayshane
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    ok thank you for the answers i will study more...

    • one year ago
  27. jayshane
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    but for now please help me just last 2 questions..!!

    • one year ago
  28. tkhunny
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    Good call. Show your work, too. This will help us as much as it helps you.

    • one year ago
  29. kirbykirby
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    Well I am not spending 1/2 my time remembering trig :P I do remember the identities but I just don't necessarily apply them right away unless I am stuck, or just force myself to be "oh maybe I should use a trig identity here". Do you know what I mean?

    • one year ago
  30. jayshane
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    my teacher encourages us to use the long method and i just cant get it,,

    • one year ago
  31. kirbykirby
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    And anyway even if you do forget, it's very easy to pull up a cal book and look for the identities, (although you should know them for the exam!!)

    • one year ago
  32. tkhunny
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    @kirbykirby Right. No ridicule intended. "1/2" was a bit hyperbolic. There are many survival methodologies. I was kind of a freak in my first differential equations class, so that does tend to make me think my survival method is the only way to go. I do realize there are humans on this planet. :-)

    • one year ago
  33. terenzreignz
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    @jayshane Your teacher recommends the long method for your practice It's not on every opportunity that knowledge of trigonometric identities will save you from long calculations, BUT the long method is the only method that would work for all situations... Now, you said you have two more questions?

    • one year ago
  34. kirbykirby
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    @tkhunny Ok good to know we are on good terms :P As for @jayshane, you still need help right

    • one year ago
  35. jayshane
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    y=tan(xsinx)

    • one year ago
  36. tkhunny
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    @jayshane I can think of only one reason to encourage a class to use the "long method". Your teacher knows your differential calculus isn't very good and is forcing you to practice! :-)

    • one year ago
  37. terenzreignz
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    Chain rule, @jayshane A good way to work on this is to identify what you could call the "outermost" function, the function that sort of "contains" all the others. Which function seems to fit this description?

    • one year ago
  38. jayshane
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    she does just give this assignment to us and she doesnt even give explanations she just gave the identities and were having a hard time to deal with it..!

    • one year ago
  39. terenzreignz
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    Well, @jayshane That's why we're here :) Now, which function is outermost?

    • one year ago
  40. jayshane
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    sin?

    • one year ago
  41. terenzreignz
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    Ahh, but sin itself is inside tangent, no?

    • one year ago
  42. kirbykirby
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    Yeah A good trick for the chain rule as @terenzreignz mentions, identity your outer functions and inner functions for the chain rule. Think of what the elementary function is and what changed: Here, we could think of tan x being the elementary function, but we changed x to "xsin x". So, you can think of it like a box: you have xsinx in the box tan[ ] , so you find the derivative of tan [ ] and multiply by the derivative inside the box [ ]

    • one year ago
  43. terenzreignz
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    y=tan(xsinx) And I've already named the outermost function... So you know tan contains all others, what's the derivative of tan?

    • one year ago
  44. jayshane
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    sec2x du/dx

    • one year ago
  45. terenzreignz
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    leave out the du/dx, and stick to sec² x So "write that down" except instead of x, copy whatever was "inside" the tan function, namely, xsin x (NOTE: We're not yet done!)

    • one year ago
  46. tkhunny
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    \(f(x) = \tan(x\cdot\sin(x))\) -- What a fascinating function. I'm not sure I've ever seen that one in a problem set.

    • one year ago
  47. jayshane
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    ok then wats nxt?

    • one year ago
  48. oldrin.bataku
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    Remember the Pythagorean identity? \(\cos^2x+\sin^2x=1\); if you divide through by \(\cos^2x\), it yields \(1+\tan^2x=\sec^2x\). We can apply this identity to simplify our function! $$y=\sec^2x-\tan^2x=\tan^2x+1-\tan^2x=1$$Do you know what the derivative of a constant function is? What is the slope of a horizontal line?

    • one year ago
  49. terenzreignz
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    Well, I like to think of chain rule as sort of a hitman's list, and think of it this way, you've already "taken 'tan' out of the picture" so you go for its "inner circle" lol anyway So now that you've "dealt with" tan (lol) time to deal with whatever was inside it. So remember sec²(xsin x) ? You multiply this to whatever the derivative of the inner function is, namely, xsin x. So what's the derivative of xsin x?

    • one year ago
  50. kirbykirby
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    @oldrin.bataku We have covered the answer for that already :) We are approaching a new question. Sorry to burst your bubble

    • one year ago
  51. jayshane
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    -cosx

    • one year ago
  52. terenzreignz
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    No... remember xsin x is the product of two functions, now would be a great time to recall your product rule?

    • one year ago
  53. kirbykirby
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    That\s not exactly right. You have x*sin x, not just "sin x" In fact you have a product rule with functions "x" and "sin x" being multiplied

    • one year ago
  54. kirbykirby
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    d/dx (x*sinx) = (1)sinx + x(cosx) by the product rule

    • one year ago
  55. oldrin.bataku
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    @kirbykirby I posted to explain the identity because none of you had. @jayshane Why don't we break our function into pieces? and *chain* the derivatives together? $$u=x\sin x\\y=\tan v\\\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=(-\sec^2u)(\sin x+x\cos x)=-\sec^2(x\sin x)[\sin x+x \cos x]$$

    • one year ago
  56. jayshane
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    (xdsinx)[sinxdx]

    • one year ago
  57. terenzreignz
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    @oldrin.bataku I think I did post the identity, though I didn't derive it. @jayshane I can see where you're getting at, for as long as dsin x is the derivative of sin x and dx is the derivative of x. What's the derivative of sin x? What's the derivative of x?

    • one year ago
  58. terenzreignz
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    Also, you don't multiply them, you add them.

    • one year ago
  59. jayshane
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    sin2x

    • one year ago
  60. terenzreignz
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    Hang on, let me rewrite your statement properly, in a way that not even your teacher can find an error in :) \[x\frac{d}{dx}\sin x +(\sin x)\frac{d}{dx}x\] Now work out these derivatives.

    • one year ago
  61. oldrin.bataku
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    @jayshane the product rule tells us \(\frac{d}{dx}uv=\frac{du}{dx}v+\frac{dv}{dx}u\) -- see how we multiply one of the original functions by the derivative of the other?.

    • one year ago
  62. jayshane
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    yes im following

    • one year ago
  63. oldrin.bataku
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    In our case of \(\frac{d}{dx}x\sin x\), we have two functions being multiplied: \(x\) and \(\sin x\). Do you know how to determine the derivative of this product using the product rule?

    • one year ago
  64. kirbykirby
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    I actually find it easier to forget the "u" notation when first learning the chain rule. Just look at it as outer and inner functions. And you won't ever need the u-notation once you master it that way

    • one year ago
  65. terenzreignz
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    @kirbykirby This is product rule now, though... Eventually we do do away with the u and v, probably

    • one year ago
  66. jayshane
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    @oldrin.bataku yeah

    • one year ago
  67. kirbykirby
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    Oops I meant the product rule LOL sorry. I am responding to another post and I am getting confused between both posts

    • one year ago
  68. jayshane
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    x(-cosx)+sinx(1)

    • one year ago
  69. terenzreignz
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    Almost... What's the derivative of sin x?

    • one year ago
  70. jayshane
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    -cos x

    • one year ago
  71. kirbykirby
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    No not quite. d/dx (sin x) = cos x d/dx (cos x ) = -sin x... yes it's confusing at first

    • one year ago
  72. terenzreignz
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    Yeah, derivative of sine is cosine Proof on request.

    • one year ago
  73. jayshane
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    can u give me appropriate solution guys, then ill just study it..

    • one year ago
  74. terenzreignz
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    Well, if you accept that the derivative of sin x is cos x, then you can work out (PLEASE DO WORK IT OUT) That the derivative of xsin x is x(cos x) + sin x Which as a whole expression, you multiply that one we got earlier, which was sec²(xsin x) Hence, you get [sec²(xsin x)][x(cos x) + sin x]

    • one year ago
  75. jayshane
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    then how will i continue to solve? the answer in the book is 4tanxsec2x

    • one year ago
  76. jayshane
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    Y=sec4x-tan4x

    • one year ago
  77. kirbykirby
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    Does your prof require that form of an answer (4tanx sec2x)?

    • one year ago
  78. jayshane
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    not exactly it is just an answer in the book.

    • one year ago
  79. kirbykirby
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    Because I can guarantee you it's the answer :) http://www.wolframalpha.com/input/?i=d%2Fdx+tan%28xsinx%29

    • one year ago
  80. jayshane
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    another question sir kirbs Y=sec4x-tan4x

    • one year ago
  81. kirbykirby
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    I tried equation your textbook answer with that one and I don't get an equality... which means the book might have a typo. Either way, it is way too much work to get to an answer like that :)

    • one year ago
  82. kirbykirby
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    You can proceed in a similar way to what I did. Or, you can make use of the identity: 1+tan^2 x = sec^2 x So:

    • one year ago
  83. kirbykirby
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    \[\sec^4x-\tan^4x = (\sec^2x)^2-\tan^4x\] can you see that?

    • one year ago
  84. jayshane
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    yes

    • one year ago
  85. kirbykirby
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    \[=(1+\tan^2x)^2-\tan^4x\] \[=(1+2\tan^2x+\tan^4x)-\tan^4x\] by using the identity (a+b)^2 = (a+2ab+b^2)

    • one year ago
  86. kirbykirby
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    \[=1+2\tan^2x\]now our problem is much simpler! We find \[\frac{d}{dx}(1+2\tan^2x)=\frac{d}{dx}1+\frac{d}{dx}2\tan^2x\]

    • one year ago
  87. kirbykirby
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    \[=0+2\frac{d}{dx}\tan^2x\] \[=2*(2*\tan x)(\sec^2 x) \] since this is the famous chain rule again. You put down the ^2 from tan( ) in front, and you multiply by the derivative of tan x which is sec^2 x

    • one year ago
  88. kirbykirby
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    \[=4\tan x \sec^2x\]

    • one year ago
  89. kirbykirby
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    Think of your box: you have \[[\tan x]^2 = 2[\tan x]*\frac{d}{dx}[\tan x]\]

    • one year ago
  90. kirbykirby
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    I hope it's clear

    • one year ago
  91. kirbykirby
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    In general for a "power" you have like \[[f(x)]^n=n[f(x)]^{n-1}*\frac{d}{dx}f(x)\] I hope I am not confusing you by writing this

    • one year ago
  92. jayshane
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    =(1+2tan2x+tan4x)−tan4x how did u get this kirb?

    • one year ago
  93. kirbykirby
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    There should be a derivative d/dx in front of [f(x)]^n, sorry my computer was lagging

    • one year ago
  94. kirbykirby
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    Ok you know that \[(a+b)^2 = (a+2ab+b^2)\]

    • one year ago
  95. jayshane
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    ahh

    • one year ago
  96. kirbykirby
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    \[(1+\tan^2x)^2=(1+2(1)(\tan^2x)+(\tan^2x)^2) \].... here we have that \[a = 1, b = \tan^2x\]

    • one year ago
  97. jayshane
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    THANK YOU A LOT KIRB...

    • one year ago
  98. kirbykirby
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    Your welcome :)

    • one year ago
  99. kirbykirby
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    Using the identity \[1+\tan^2x=\sec^2x\]We have \[\sec^2x-\tan^2x=(1+\tan^2x)-\tan^2x\] \[=1\] So, \[\frac{d}{dx}1=0\]

    • one year ago
  100. jayshane
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    TY KIRB

    • one year ago
  101. kirbykirby
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    :)

    • one year ago
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