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jayshane

  • one year ago

how about differentiate: y=sec²x-tan²x

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  1. terenzreignz
    • one year ago
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    Derivative of the sum is the sum of the derivatives.

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

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

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

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

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

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

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

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

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

  11. terenzreignz
    • one year ago
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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...?

  12. kirbykirby
    • one year ago
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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

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

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

  15. tkhunny
    • one year ago
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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!

  16. kirbykirby
    • one year ago
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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 :)

  17. kirbykirby
    • one year ago
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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

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

  19. tkhunny
    • one year ago
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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.

  20. kirbykirby
    • one year ago
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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

  21. jayshane
    • one year ago
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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...

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

  23. kirbykirby
    • one year ago
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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 :)

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

  25. tkhunny
    • one year ago
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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?

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

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

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

  29. kirbykirby
    • one year ago
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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?

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

  31. kirbykirby
    • one year ago
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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!!)

  32. tkhunny
    • one year ago
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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. :-)

  33. terenzreignz
    • one year ago
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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?

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

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

  36. tkhunny
    • one year ago
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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! :-)

  37. terenzreignz
    • one year ago
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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?

  38. jayshane
    • one year ago
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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..!

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

  40. jayshane
    • one year ago
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    sin?

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

  42. kirbykirby
    • one year ago
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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 [ ]

  43. terenzreignz
    • one year ago
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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?

  44. jayshane
    • one year ago
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    sec2x du/dx

  45. terenzreignz
    • one year ago
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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!)

  46. tkhunny
    • one year ago
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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.

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

  48. oldrin.bataku
    • one year ago
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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?

  49. terenzreignz
    • one year ago
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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?

  50. kirbykirby
    • one year ago
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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

  51. jayshane
    • one year ago
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    -cosx

  52. terenzreignz
    • one year ago
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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?

  53. kirbykirby
    • one year ago
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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

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

  55. oldrin.bataku
    • one year ago
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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? $$u=x\sin x\\y=\tan v\\\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=(-\sec^2u)(\sin x+x\cos x)=-\sec^2(x\sin x)[\sin x+x \cos x]$$

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

  57. terenzreignz
    • one year ago
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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?

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

  59. jayshane
    • one year ago
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    sin2x

  60. terenzreignz
    • one year ago
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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.

  61. oldrin.bataku
    • one year ago
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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?.

  62. jayshane
    • one year ago
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    yes im following

  63. oldrin.bataku
    • one year ago
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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?

  64. kirbykirby
    • one year ago
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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

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

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

  67. kirbykirby
    • one year ago
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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

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

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

  70. jayshane
    • one year ago
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    -cos x

  71. kirbykirby
    • one year ago
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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

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

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

  74. terenzreignz
    • one year ago
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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]

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

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

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

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

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

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

  81. kirbykirby
    • one year ago
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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 :)

  82. kirbykirby
    • one year ago
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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:

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

  84. jayshane
    • one year ago
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    yes

  85. kirbykirby
    • one year ago
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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)

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

  87. kirbykirby
    • one year ago
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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

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

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

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

  91. kirbykirby
    • one year ago
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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

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

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

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

  95. jayshane
    • one year ago
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    ahh

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

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

  98. kirbykirby
    • one year ago
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    Your welcome :)

  99. kirbykirby
    • one year ago
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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, \[\frac{d}{dx}1=0\]

  100. jayshane
    • one year ago
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    TY KIRB

  101. kirbykirby
    • one year ago
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    :)

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